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
4 years, 9 months 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

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

- 4 years, 9 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 equation

- 4 years, 9 months ago

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.

- 4 years, 9 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.

- 4 years, 9 months ago

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

- 4 years, 9 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.

- 4 years, 9 months ago

thank you so much sir for the help $\ddot \smile$

- 4 years, 9 months ago

but sir shouldn't it be $v\frac{dv}{dx}+e^{2y}v^{3}=0$

- 4 years, 9 months ago

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. :)

- 4 years, 9 months ago

- 4 years, 9 months ago