partial differential equation - 11

logistic_guy

Senior Member
Joined
Apr 17, 2024
Messages
2,214
\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle ax^2\frac{\partial w}{\partial x} - bx^2\frac{\partial w}{\partial y} = (w - bx - ay)^2\)
 
\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle ax^2\frac{\partial w}{\partial x} - bx^2\frac{\partial w}{\partial y} = (w - bx - ay)^2\)
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem
 
Beat it.

\(\displaystyle \frac{dx}{ax^2} = \frac{dy}{-bx^2} = \frac{dw}{(w - bx - ay)^2}\)


\(\displaystyle \frac{dx}{ax^2} = \frac{dy}{-bx^2}\)


\(\displaystyle \frac{dx}{a} = \frac{dy}{-b}\)


\(\displaystyle b \ dx = -a \ dy\)


\(\displaystyle b \int dx = -a \int dy\)


\(\displaystyle bx = -ay + C_1\)


\(\displaystyle bx + ay = C_1\)

Then,

\(\displaystyle \Phi(C_1) = \Phi(bx + ay) = C_2\)
 
The next steps are very hard to solve.

We start with the first coefficient.

\(\displaystyle \frac{dx}{ds} = ax^2\)


\(\displaystyle \frac{dx}{x^2} = a \ ds\)


\(\displaystyle \int\frac{dx}{x^2} = \int a \ ds\)


\(\displaystyle -\frac{1}{x} = as + E\)
 
It's \(\displaystyle \textcolor{black}{\bold{Hard}}\)😭😭

From the previous post, we have:

\(\displaystyle x = -\frac{1}{as + E}\)

Now we try to crack the second coefficient.

\(\displaystyle \frac{dy}{ds} = -bx^2\)


\(\displaystyle dy = -bx^2 \ ds\)


\(\displaystyle \int dy = -b\int x^2 \ ds\)


\(\displaystyle \int dy = -b\int \frac{1}{(as + E)^2} \ ds\)


\(\displaystyle y = \frac{b}{a}\left(\frac{1}{as + E}\right) + F\)
 
Live the moment.

\(\displaystyle \frac{dw}{ds} = (w - bx - ay)^2\)


Let \(\displaystyle z = w - bx - ay\).


\(\displaystyle \frac{dz}{ds} = \frac{dw}{ds} - b\frac{dx}{ds} - a\frac{dy}{ds} = (w - bx - ay)^2 - abx^2 + abx^2 = z^2\)


\(\displaystyle \int \frac{dz}{z^2} = \int ds\)


\(\displaystyle -\frac{1}{z} = s + G\)


\(\displaystyle z = -\frac{1}{s + G}\)


\(\displaystyle w - bx - ay = -\frac{1}{-\frac{1}{ax} - \frac{E}{a} + G}\)


\(\displaystyle w - bx - ay = \frac{1}{\frac{1}{ax} - G + \frac{E}{a}}\)


\(\displaystyle w - bx - ay = \frac{1}{\frac{1}{ax} - H}\)


\(\displaystyle w - bx - ay = \frac{ax}{1 - Hax}\)


\(\displaystyle w = \frac{ax}{1 - Hax} + bx + ay\)


\(\displaystyle w = \frac{ax + (bx + ay)(1 - Hax)}{1 - Hax}\)


\(\displaystyle w = \frac{ax + bx + ay - (bx + ay)Hax}{1 - Hax}\)


\(\displaystyle w = \frac{(a + b)x + ay - (bx + ay)Hax}{1 - Hax}\)


\(\displaystyle w = \frac{(a + b)x + ay - \frac{(bx + ay)ax}{C_2}}{1 - \frac{ax}{C_2}}\)


\(\displaystyle w = \frac{(a + b)xC_2 + ayC_2 - (bx + ay)ax}{C_2 - ax}\)


\(\displaystyle w = \frac{\bigg[(a + b)x + ay\bigg]C_2 - (bx + ay)ax}{C_2 - ax}\)

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

\(\displaystyle w = \frac{\bigg[(a + b)x + ay\bigg]\Phi(bx + ay) - (bx + ay)ax}{\Phi(bx + ay) - ax}\)
 
Top