Miscellaneous Problems. One of the difficulties in solving first order

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = x3 2y x

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 1 + cos x 2 sin y

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 2x + y 3 + 3y2 x , y(0) = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 3 6x + y 2xy

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 2xy + y2 + 1 x2 + 2xy

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. x dy dx + xy = 1 y, y(1) = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 4x3 + 1 y(2 + 3y)

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. x dy dx + 2y = sin x x , y(2) = 1

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 2xy + 1 x2 + 2y

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. (x2y + xy y) + (x2y 2x2) dy dx = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. . (x2 + y) + (x + ey) dy dx = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx + y = 1 1 + ex

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 1 + 2x + y2 + 2xy2

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. (x + y) + (x + 2y) dy dx = 0, y(2) = 3

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. (ex + 1) dy dx = y yex

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = ex cos y e2y cos x ex sin y + 2e2y sin x

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = e2x + 3y

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx + 2y = ex22x, y(0) = 3

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 3x2 2y y3 2x + 3xy2

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. y= ex+y

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx + 2y2 + 6xy 4 3x2 + 4xy + 3y2 = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = x2 1 y2 + 1 , y(1) = 1

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. t dy dt + (t + 1)y = e2t

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. 2 sin y sin x cos x + cos y sin2 x dy dx = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.  2 x y y x2 + y2 + x x2 + y2 x2 y2 dy dx = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. xy= y + xey/x

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = x x2y + y3 Hint: Let u = x2

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. (2y + 3x) = x dy dx

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = x + y x y

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. (3y2 + 2xy) (2xy + x2) dy dx = 0

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. dy dx = 3x2y + y2 2x3 + 3xy , y(1) = 2

Miscellaneous Problems. One of the difficulties in solving first order equations is that there are several methods of solution, each of which can be used on a certain type of equation. It may take some time to become proficient in matching solution methods with equations. The first 32 of the following problems are presented to give you some practice in identifying the method or methods applicable to a given equation. The remaining problems involve certain types of equations that can be solved by specialized methods. In each of Problems 1 through 32, solve the given differential equation. If an initial condition is given, also find the solution that satisfies it. xy+ y y2e2x = 0, y(1) = 2

Riccati Equations. The equation dy dt = q1(t) + q2(t)y + q3(t)y2 is known as a Riccati23 equation. Suppose that some particular solution y1 of this equation is known. A more general solution containing one arbitrary constant can be obtained through the substitution y = y1(t) + 1 v(t) . Show that v(t) satisfies the first order linear equation dv dt = (q2 + 2q3y1)v q3. Note that v(t) will contain a single arbitrary constant.

Using the method of Problem 33 and the given particular solution, solve each of the following Riccati equations: (a) y= 1 + t 2 2ty + y2; y1(t) = t (b) y= 1 t2 y t + y2; y1(t) = 1 t (c) dy dt = 2 cos2 t sin2 t + y2 2 cost ; y1(t) = sin t

The propagation of a single action in a large population (for example, drivers turning on headlights at sunset) often depends partly on external circumstances (gathering darkness) and partly on a tendency to imitate others who have already performed the action in question. In this case the proportion y(t) of people who have performed the action can be described24 by the equation dy/dt = (1 y)[x(t) + by], (i) where x(t) measures the external stimulus and b is the imitation coefficient. (a) Observe that Eq. (i) is a Riccati equation and that y1(t) = 1 is one solution. Use the transformation suggested in Problem 33, and find the linear equation satisfied by v(t). (b) Find v(t) in the case that x(t) = at, where a is a constant. Leave your answer in the form of an integral.

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. t 2y+ 2ty 1 = 0, t > 0

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. ty+ y= 1, t > 0

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. y+ t(y )2 = 0

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. 2t 2y+ (y )3 = 2ty , t > 0

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. y+ y= et

Some Special Second Order Equations. Second order equations involve the second derivative of the unknown function and have the general form y= f(t, y, y ). Usually such equations cannot be solved by methods designed for first order equations. However, there are two types of second order equations that can be transformed into first order equations by a suitable change of variable. The resulting equation can sometimes be solved by the methods presented in this chapter. Problems 36 through 51 deal with these types of equations. Equations with the Dependent Variable Missing. For a second order differential equation of the form y= f(t, y ), the substitution v = y , v= yleads to a first order equation of the form v= f(t, v). If this equation can be solved for v, then y can be obtained by integrating dy/dt = v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y. In each of Problems 36 through 41, use this substitution to solve the given equation. t 2y= (y )2, t > 0

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. yy+ (y )2 = 0

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. y+ y = 0

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. y+ y(y )3 = 0

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. 2y2y+ 2y(y )2 = 1

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. yy (y )3 = 0

Equations with the Independent Variable Missing. Consider second order differential equations of the form y= f(y, y ), in which the independent variable t does not appear explicitly. If we let v = y , then we obtain dv/dt = f(y, v). Since the right side of this equation depends on y and v, rather than on t and v, this equation contains too many variables. However, if we think of y as the independent variable, then by the chain rule, dv/dt = (dv/dy)(dy/dt) = v(dv/dy). Hence the original differential equation can be written as v(dv/dy) = f(y, v). Provided that this first order equation can be solved, we obtain v as a function of y. A relation between y and t results from solving dy/dt = v(y), which is a separable equation. Again, there are two arbitrary constants in the final result. In each of Problems 42 through 47, use this method to solve the given differential equation. y+ (y )2 = 2ey Hint: In Problem 47 the transformed equation is a Bernoulli equation. See Problem 27 in Section 2.4.

In each of Problems 48 through 51, solve the given initial value problem using the methods of Problems 36 through 47. y y= 2, y(0) = 1, y (0) = 2

In each of Problems 48 through 51, solve the given initial value problem using the methods of Problems 36 through 47. y 3y2 = 0, y(0) = 2, y (0) = 4

In each of Problems 48 through 51, solve the given initial value problem using the methods of Problems 36 through 47. (1 + t 2)y+ 2ty+ 3t 2 = 0, y(1) = 2, y (1) = 1

In each of Problems 48 through 51, solve the given initial value problem using the methods of Problems 36 through 47. y y t = 0, y(1) = 2, y (1) = 1