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I need help in solving this problem

Find the general solution of the partial differential equation $$x(z+2a)p+(xz+2yz+2ay)q=z(z+a),$$ where $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$. Also find its particular integral surface which passes through the given curve $y=0$, $z^2=4ax$.

I know this is a Lagrange's linear equation and that I require two general solution of this equation to find the Integral surface.

I got the first general solution like this $$\frac{dx}{x(z+2a)}=\frac{dy}{xz+2yz+2ay}=\frac{dz}{z(z+a)}$$ comparing 1st and 3rd we get $$\frac{dx}{x}=\frac{(z+2a)dz}{z(z+a)}$$ Integrating this we get the first solution $$\frac{z^2}{x(z+a)}=c_1.$$ I just can not find the second solution , I have tried substituting $$x=\frac{z^2}{c_1(z+a)}$$ while equating the 2nd and 3rd equation but it gets much more complex.

Can you find the second solution?

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    $\begingroup$ Using Componendo-Dividendo on the first equality yields \begin{align} \frac{dx + dy}{2xz + 2yz + 2ax + 2ay} &= \frac{d (x + y)}{(2a + 2z)(x + y)} \quad \text {(C.D)}\\ &= \frac{dz}{z(z + a)} \\ \implies \frac{d (x + y)}{x + y} &= \frac{2 dz}{z} \\ \implies \ln(x + y) &= \ln(z^{2}) + c_{1} \\ \implies \frac{x + y}{z^{2}} &= c_{1} \end{align} $\endgroup$ – mattos Apr 27 '18 at 12:43

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