Couple of doubts

I am facing problem in solving these differential equation

1) d2ydx2+e2y(dydx)3=0\large{\frac{d^{2}y}{dx^{2}}+e^{2y}(\frac{dy}{dx})^{3}=0}

2)d2ydx2+3y=2x\large{\frac{d^{2}y}{dx^{2}}+3y=-2x}

Note by Tanishq Varshney
4 years, 3 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 ±i3,\pm i\sqrt{3}, so yc=c1cos(3x)+c2sin(3x).y_{c} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x).

For the particular solution, letting yp=2x3y_{p} = -\dfrac{2x}{3} will do the trick, and thus the general solution is

y=yc+yp=c1cos(3x)+c2sin(3x)2x3.y = y_{c} + y_{p} = c_{1}\cos(\sqrt{3}x) + c_{2}\sin(\sqrt{3}x) - \dfrac{2x}{3}.

Brian Charlesworth - 4 years, 3 months ago

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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 - 4 years, 3 months ago

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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=dydx.v = \dfrac{dy}{dx}. The equation then becomes

vdvdy+e2yv3=0,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 - 4 years, 3 months ago

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@Brian Charlesworth For the first one of the terms on the other side and divide by (dydx)3\left(\frac{dy}{dx}\right)^3 on both the sides to get the equation of the form d2xdy2=e2y\displaystyle \frac{d^{2}x}{dy ^{2}} = - e^{2y} which is then trivial.

Sudeep Salgia - 4 years, 3 months ago

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@Sudeep Salgia sorry not understood, how is d2ydx2(dydx)3=d2xdy2\huge{\frac{\frac{d^{2}y}{dx^{2}}}{(\frac{dy}{dx})^{3}}=\frac{d^{2}x}{dy^{2}}}

Tanishq Varshney - 4 years, 3 months ago

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@Tanishq Varshney It is simple. Start with (dxdy)=(dydx)1\left(\frac{dx}{dy }\right) = \left(\frac{dy}{dx}\right)^{-1} . Then differentiate wrt yy and apply chain rule on the other side to get the required answer.

Sudeep Salgia - 4 years, 3 months ago

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@Brian Charlesworth thank you so much sir for the help ¨\ddot \smile

Tanishq Varshney - 4 years, 3 months ago

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@Brian Charlesworth but sir shouldn't it be vdvdx+e2yv3=0v\frac{dv}{dx}+e^{2y}v^{3}=0

Tanishq Varshney - 4 years, 3 months ago

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@Tanishq Varshney With dydx=v\dfrac{dy}{dx} = v we have, by way of the chain rule, that d2ydx2=dvdx=dvdydydx=vdvdy.\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 xx as the independent variable into a first-order DE with yy as the independent variable. We can simplify a bit further by eliminating v=0v = 0 as a solution, giving us

dvdy+e2yv2=0,\dfrac{dv}{dy} + e^{2y}v^{2} = 0, which is separable with solution v=2e2y+C.v = \dfrac{2}{e^{2y} + C}.

So now we have that dydx=2e2y+C,\dfrac{dy}{dx} = \dfrac{2}{e^{2y} + C}, which again is separable with solution e2y2+Cy=2x+K,\dfrac{e^{2y}}{2} + Cy = 2x + K,

where constants CC and KK would be determined from any initial conditions that might be provided. We can't form a "nice" equation in the form y=f(x)y = f(x) from this equation, but we've come far enough. :)

Brian Charlesworth - 4 years, 3 months ago

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