Find two solutions of the initial-value problem Use a numerical solver to graph the solution curves.

Solution :Step 1 of 5 : In this problem, we need to find the two solutions of the initial-value problem Step 2 of 5 : Find the solution of equation preview Let u = y’u’ = y’’The equation becomes (u’’)² + u² = 1 (u’’)² + u² -1 = 0Solve this quadratic equation We get equation preview Separate the variables and solve this two equations Step 3 of 5 :Let us take the positive u’ equation preview equation preview Take the integral and simplify this equation preview equation preview equation preview We know that u = y’ equation preview Apply the initial conditions y’(/2)=3/2 and solve for equation preview We get = /6 equation preview equation preview equation preview equation preview Again plug the first initial condition y(/2)=½ and solve for equation preview We get = 1 The solution is equation preview