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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
1 year, 10 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

$$y = y_{c} + y_{p} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x) - \dfrac{2x}{3}.$$ · 1 year, 10 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 · 1 year, 10 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. · 1 year, 10 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. · 1 year, 10 months ago

sorry not understood, how is $$\huge{\frac{\frac{d^{2}y}{dx^{2}}}{(\frac{dy}{dx})^{3}}=\frac{d^{2}x}{dy^{2}}}$$ · 1 year, 10 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. · 1 year, 10 months ago

but sir shouldn't it be $$v\frac{dv}{dx}+e^{2y}v^{3}=0$$ · 1 year, 10 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. :) · 1 year, 10 months ago

thank you so much sir for the help $$\ddot \smile$$ · 1 year, 10 months ago