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

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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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

\(y = y_{c} + y_{p} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x) - \dfrac{2x}{3}.\) – Brian Charlesworth · 1 year, 5 months ago

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– Tanishq Varshney · 1 year, 5 months ago

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 equationLog in to reply

link, just to save some time. The headings "Complex Roots" and "Nonhomogeneous DE's" are applicable for 2).

I'll provide you with a greatFor 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 · 1 year, 5 months ago

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– Sudeep Salgia · 1 year, 5 months ago

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.Log in to reply

– Tanishq Varshney · 1 year, 5 months ago

sorry not understood, how is \(\huge{\frac{\frac{d^{2}y}{dx^{2}}}{(\frac{dy}{dx})^{3}}=\frac{d^{2}x}{dy^{2}}}\)Log in to reply

– Sudeep Salgia · 1 year, 5 months ago

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.Log in to reply

– Tanishq Varshney · 1 year, 5 months ago

but sir shouldn't it be \(v\frac{dv}{dx}+e^{2y}v^{3}=0\)Log in to reply

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 · 1 year, 5 months ago

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– Tanishq Varshney · 1 year, 5 months ago

thank you so much sir for the help \(\ddot \smile\)Log in to reply

@Caleb Townsend @Ronak Agarwal @Otto Bretscher sir – Tanishq Varshney · 1 year, 5 months ago

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@Brian Charlesworth sir, @Raghav Vaidyanathan ,@Shashwat Shukla – Tanishq Varshney · 1 year, 5 months ago

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