I am facing problem in solving these differential equation

1) \(\large{\frac{d^{2}y}{dx^{2}}+e^{2y}(\frac{dy}{dx})^{3}=0}\)

2)\(\large{\frac{d^{2}y}{dx^{2}}+3y=-2x}\)

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## Comments

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TopNewest1) is non-linear, so is going to be messy. I'll have a look at 2) though.

The characteristic equation has roots \(\pm i\sqrt{3},\) so \(y_{c} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x).\)

For the particular solution, letting \(y_{p} = -\dfrac{2x}{3}\) will do the trick, and thus the general solution is

\(y = y_{c} + y_{p} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x) - \dfrac{2x}{3}.\)

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sir can u provide me the steps, or any note on how to solve these, actually i only know how to solve homogeneous second order differential equation

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I'll provide you with a great link, just to save some time. The headings "Complex Roots" and "Nonhomogeneous DE's" are applicable for 2).

For 1) you could try the substitution \(v = \dfrac{dy}{dx}.\) The equation then becomes

\(v\dfrac{dv}{dy} + e^{2y}v^{3} = 0,\) which is still non-linear but is now first-order and may be easier to solve.

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So we've converted the original second-order DE with \(x\) as the independent variable into a first-order DE with \(y\) as the independent variable. We can simplify a bit further by eliminating \(v = 0\) as a solution, giving us

\(\dfrac{dv}{dy} + e^{2y}v^{2} = 0,\) which is separable with solution \(v = \dfrac{2}{e^{2y} + C}.\)

So now we have that \(\dfrac{dy}{dx} = \dfrac{2}{e^{2y} + C},\) which again is separable with solution \(\dfrac{e^{2y}}{2} + Cy = 2x + K,\)

where constants \(C\) and \(K\) would be determined from any initial conditions that might be provided. We can't form a "nice" equation in the form \(y = f(x)\) from this equation, but we've come far enough. :)

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@Brian Charlesworth sir, @Raghav Vaidyanathan ,@Shashwat Shukla

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@Caleb Townsend @Ronak Agarwal @Otto Bretscher sir

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