Why we would get one of the solution of $\displaystyle\frac{dy}{dx}=\frac{2y}{x}+\cos(\frac{y}{x^2})$ as $y=[\displaystyle\frac{\pi}{2}+(2k+1)\pi]x^2$
How to find out the solution of the differential equation?
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ordinary-differential-equations
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0is there such tech$n$ique to solve ODE? – 2012-02-22
1 Answers
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Let us put: $y=v(x).x^2$ (I hope this is an elementary intuition, which comes at first glance)
Then your ODE becomes $x^2\frac{dv}{dx}=\cos v$ Which has the solution $\tan \left(\frac{\pi}{4}+\frac{v}{2}\right)=e^{-\frac{1}{x}+c}$. To get your desired solution (which I should never think of, unless being asked) make both sides zero by taking $c=-\infty$.
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0All I can say that $c$ is an arbitrary constant.... – 2012-02-23