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Couple of doubts

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

Note by Tanishq Varshney
2 years ago

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1) 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

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

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@Brian Charlesworth 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 Tanishq Varshney · 2 years ago

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@Tanishq Varshney 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. Brian Charlesworth · 2 years ago

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@Brian Charlesworth For the first one of the terms on the other side and divide by \(\left(\frac{dy}{dx}\right)^3\) on both the sides to get the equation of the form \(\displaystyle \frac{d^{2}x}{dy ^{2}} = - e^{2y}\) which is then trivial. Sudeep Salgia · 2 years ago

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@Sudeep Salgia sorry not understood, how is \(\huge{\frac{\frac{d^{2}y}{dx^{2}}}{(\frac{dy}{dx})^{3}}=\frac{d^{2}x}{dy^{2}}}\) Tanishq Varshney · 2 years ago

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@Tanishq Varshney It is simple. Start with \(\left(\frac{dx}{dy }\right) = \left(\frac{dy}{dx}\right)^{-1} \). Then differentiate wrt \(y\) and apply chain rule on the other side to get the required answer. Sudeep Salgia · 2 years ago

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@Brian Charlesworth but sir shouldn't it be \(v\frac{dv}{dx}+e^{2y}v^{3}=0\) Tanishq Varshney · 2 years ago

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@Tanishq Varshney With \(\dfrac{dy}{dx} = v\) we have, by way of the chain rule, that \(\dfrac{d^{2}y}{dx^{2}} = \dfrac{dv}{dx} = \dfrac{dv}{dy}\dfrac{dy}{dx} = v\dfrac{dv}{dy}.\)

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. :) Brian Charlesworth · 2 years ago

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@Brian Charlesworth thank you so much sir for the help \(\ddot \smile\) Tanishq Varshney · 2 years ago

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