Mimi_x3
  • Mimi_x3
https://gyazo.com/63b9663673768d74accfa80d12f95ae7
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mimi_x3
  • Mimi_x3
help pls <3
Loser66
  • Loser66
Verify part: y= sin (x^2) , hence y ' = 2x cos (x^2), y"= 2cos(x^2) -4x^2 sin(x^2) Plug back \(x(2cos (x^2) - 4x^2 sin(x^2) -2x cos(x^2) +4x^3 sin(x^2) =0\)
Loser66
  • 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 \(y_2\)

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