partial differential equation - 9

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = w - a\sqrt{w^2 - x^2 - y^2}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = w - a\sqrt{w^2 - x^2 - y^2}\)

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Please share your work/thoughts about this problem

 

[logistic_guy]

The problems are getting too dense.

\(\displaystyle \frac{dx}{x} = \frac{dy}{y} = \frac{dw}{w - a\sqrt{w^2 - x^2 - y^2}}\)


\(\displaystyle \frac{dx}{x} = \frac{dy}{y}\)


\(\displaystyle \int \frac{dx}{x} = \int \frac{dy}{y}\)


\(\displaystyle \ln x + D = \ln y\)


\(\displaystyle \ln y - \ln x = D\)


\(\displaystyle \ln\frac{y}{x} = D\)


\(\displaystyle \frac{y}{x} = e^D = C_1\)

Then,

\(\displaystyle \Phi(C_1) = \Phi\left(\frac{y}{x}\right) = C_2\)

 

[logistic_guy]

The next part is very hard to solve.

\(\displaystyle \frac{dx}{x} = \frac{dw}{w - a\sqrt{w^2 - x^2 - y^2}}\)

We have to find a way to get rid of the square root. Let us try this substitution:

\(\displaystyle z = w + \sqrt{w^2 - x^2 - y^2}\)


\(\displaystyle \frac{\partial z}{\partial x} = \frac{\partial w}{\partial x} + \frac{2w\frac{\partial w}{\partial x} - 2x}{2\sqrt{w^2 - x^2 - y^2}}\)


\(\displaystyle \frac{\partial z}{\partial y} = \frac{\partial w}{\partial y} + \frac{2w\frac{\partial w}{\partial x} - 2y}{2\sqrt{w^2 - x^2 - y^2}}\)

Substitute this result in the original partial differential equation.

\(\displaystyle x\left(\frac{\partial z}{\partial x} - \frac{w\frac{\partial w}{\partial x} - x}{\sqrt{w^2 - x^2 - y^2}}\right) + y\left(\frac{\partial z}{\partial y} - \frac{w\frac{\partial w}{\partial y} - y}{\sqrt{w^2 - x^2 - y^2}}\right) = z - \sqrt{w^2 - x^2 - y^2} - a\sqrt{w^2 - x^2 - y^2}\)

Let \(\displaystyle R = \sqrt{w^2 - x^2 - y^2}\).

\(\displaystyle x\left(\frac{\partial z}{\partial x} - \frac{w\frac{\partial w}{\partial x} - x}{R}\right) + y\left(\frac{\partial z}{\partial y} - \frac{w\frac{\partial w}{\partial y} - y}{R}\right) = z - R - aR\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - R - aR + \frac{1}{R} \left( x w \frac{\partial w}{\partial x} + y w \frac{\partial w}{\partial y} - x^2 - y^2 \right)\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - R - aR + \frac{1}{R} \left( w[w -aR] - x^2 - y^2 \right)\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - R - aR + \frac{1}{R} \left( w^2 -aRw - x^2 - y^2 \right)\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - R - aR + \frac{1}{R} \left(R^2 -aRw\right)\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - R - aR + R - aw\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - aR - aw\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - aR - a(z - R)\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - aR - az + aR\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z - az\)


\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z(1 - a)\)

 

[logistic_guy]

\(\displaystyle x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}= z(1 - a)\)

We continue.

\(\displaystyle \frac{dx}{x} = \frac{dz}{z(1 - a)}\)


\(\displaystyle (1 - a)\int\frac{dx}{x} = \int\frac{dz}{z}\)


\(\displaystyle (1 - a)\ln x + E = \ln z\)


\(\displaystyle \ln x^{1 - a} + E = \ln z\)


\(\displaystyle C_2x^{1 - a} = z\)


\(\displaystyle z = \frac{C_2}{x^{a - 1}}\)

Then, the general solution to the OP partial differential equation is:

\(\displaystyle w + \sqrt{w^2 - x^2 - y^2} = \frac{1}{x^{a - 1}}\Phi\left(\frac{y}{x}\right)\)