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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

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

Log in to reply

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

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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. :)

Log in to reply

Log in to reply

@Caleb Townsend @Ronak Agarwal @Otto Bretscher sir

Log in to reply

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

Log in to reply