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Loser66
 one year ago
Best ResponseYou've already chosen the best response.2Verify part: y= sin (x^2) , hence y ' = 2x cos (x^2), y"= 2cos(x^2) 4x^2 sin(x^2) Plug back \(x(2cos (x^2)  4x^2 sin(x^2) 2x cos(x^2) +4x^3 sin(x^2) =0\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.2Reduction 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 \(y_2\)
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