Question

1. PT 24
5. xyp + y (2x - y)q = 2xz.
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Answer 1

Lagrange's Solution of a Linear Partial Differential Equation

Solution:

Here, $xyp+y(2x-y)q=2xz$

Then Lagrange's subsidiary equations are

$\quad\frac{dx}{xy}=\frac{dy}{y(2x-y)}=\frac{dz}{2xz}$

Taking the first two ratios, we get

$\quad\frac{dx}{x}=\frac{dy}{2x-y}$

$\Rightarrow 2x\:dx-y\:dx=x\:dy$

$\Rightarrow 2x\:dx-(y\:dx+x\:dy)=0$

$\Rightarrow 2x\:dx-d(xy)=0$

Taking integration on both sides, we get

$\quad \int 2x\:dx-\int d(xy)=c$ where c = int. const.

$\Rightarrow x^{2}-xy=c$ ...(1)

Now taking the first and the third ratio, we get

$\quad \frac{dx}{y}=\frac{dz}{2z}$

$\Rightarrow \frac{dx}{\frac{x^{2}-c}{c}}=\frac{dz}{2z}$ [by (1)]

$\Rightarrow\frac{2x\:dx}{x^{2}-c}-\frac{dz}{z}=0$

On integration, we get

$\quad\int\frac{2x\:dx}{x^{2}-c}-\int\frac{dz}{z}=c'$ where c' = int. const.

$\Rightarrow log(x^{2}-c)-log(z)=log(c')$

$\Rightarrow log(\frac{x^{2}-x^{2}+xy}{z}=log(c')$ [by (1)

$\Rightarrow \frac{xy}{z}=c'$ ...(2)

Hence the general solution is

$\quad \phi(x^{2}-xy,\:\frac{xy}{z})=0$ where $\phi$ is arbitrary.

Answer: $\phi(x^{2}-xy,\:\frac{xy}{z})=0$ where $\phi$ is arbitrary.

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Solve: $xyp + y (2x - y)q = 2xz$

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