## Mimi_x3 one year ago https://gyazo.com/63b9663673768d74accfa80d12f95ae7

1. Mimi_x3

help pls <3

2. Loser66

Verify part: y= sin (x^2) , hence y ' = 2x cos (x^2), y"= 2cos(x^2) -4x^2 sin(x^2) Plug back $$x(2cos (x^2) - 4x^2 sin(x^2) -2x cos(x^2) +4x^3 sin(x^2) =0$$

3. Loser66

Reduction Order: Let the second solution is $$y_2 = V(t) y_1(t)$$ , then we have (this is formula, but if you want, you can take $$y_2 ', y_2"$$ and plug to the original one to get the form, $$y_1V" + (2y_1' +y_1)V'=0$$ That is $$sin(x^2) V" +(2xcos(x^2) + sin(x^2))V' =0$$ Now let $$W = V'\rightarrow W'= V" (1)$$ We have $$sin(x^2) W' + (2xcox (x^2) + sin(x^2) W =0$$ $$\dfrac{dW}{W} = \dfrac{-(2xcos (x^2) +sin(x^2)}{sin(x^2)}dx$$ Now take integral both sides to get W, then plug back to (1) to find V' Tha is V', NOT V, hence, you have to take integral again to get V, then plug back to $$y_2 = y_1 V(t)$$ to get $$y_2$$