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I took my examination in differential equations earlier; i would've gotten a perfect score, but i tripped in this problem:

$$(1+5y\sin x)dy + y^4\cos xdx = 0$$

I used substitution suggested by the equation: $w = \sin x$, $dw = \cos x dx$. I transformed it into a linear D.E., however, i was stuck in the integration of the right part:

$$\int -y^{-4}e^{-5/2y^{-2}}dy$$

i tried using integration by parts, but i used substitution to simplify the function: $u = y^{-1}$, $du = -y^{-2}dy$, resulting in $$\int [u^2e^{-5/2u^2}du]. $$

i hoped to get a repeating integral, but the sign i got for the integral have the same sign, making it cancel with each other.

is my method wrong, or are there any method to evaluate this differential equation?

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    $\begingroup$ Please use MathJax to format your equations. I've edited the post, but you should check if it is right. $\endgroup$
    – Andrei
    Apr 18 at 13:29
  • $\begingroup$ Your method is fine. Possibly your instructor intended to check your understanding of the fundamental theorem of calculus to write the solution in terms of a definite integral,$$e^{\tfrac5{2y^2}} \sin (x(y)) = C - \int_{y_0}^y e^{\tfrac5{2t^2}}\,\frac{dt}{t^4}$$ $\endgroup$
    – user170231
    Apr 18 at 14:46
  • $\begingroup$ Wow, can you explain it a little further? I want to understand on how I can write my solution in that way. $\endgroup$ Apr 18 at 14:52
  • $\begingroup$ It's just applying the FTC:$$f'(y)=g(y) \implies f(y) - f(y_0) = \int_{y_0}^y g(t)\,dt$$Here $C$ plays the role of $f(y_0)$, and $y_0\neq0$ is an arbitrarily chosen constant. $\endgroup$
    – user170231
    Apr 18 at 15:09

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You got rightfully stuck on that integral because you can't integrate it, it can only be solved numericallly. There must have been a typo in the question. My guess is that the question should have been: $$ (1+5y^2\sin x)dy+y^4\cos xdx=0. $$ Which would have the solution: $$ \dfrac{e^\frac{-5}{y}(2y^2+10y+25)}{125y^2}+e^\frac{-5}{y}\sin x=c. $$

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  • $\begingroup$ I hope so, because I know that the question was inspired from a book I read before, which is (1+3xsiny)dx$ - x^2cosy dy$ = 0 $\endgroup$ Apr 18 at 14:26

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