2nd order partial differential equation

[pk415]

Hello all, this is my first post and I'm having trouble with some homework. Here is the problem:

Solve:
$$U_x_y - yU_y = e^x$$

I tried subbing $$V = U_y$$ then I have

$$V_x - yV = e^x$$

I solve this as a linear equation with an integrating factor of $$e^{-\frac{1}{2}y^2}$$

and get

$$V = e^{\frac{1}{2}y^2}*(e^{-\frac{1}{2}y^2} \int e^x dx + f(y))$$

$$V = e^x + e^{\frac{1}{2}y^2}*f(y)$$

or
$$U_y = e^x + e^{\frac{1}{2}y^2}*f(y)$$

Now, how do I integrate the second part wrt y?

Thanks

 

[Mute]

You have the wrong integrating factor. You integrated with respect to y to find it when you should have integrated with respect to x, treating y as a constant.

The function of y that you end up with in your result cannot be integrated without some sort of initial condition to let you determine what the function is.

For example, say you were given $U_y(x=0,y) = y$, then solving for $f(y)$ (using the equation for U_y I arrived at) gives $f(y) = y - \frac{1}{1-y}$.

An initial condition $U(x=a,y) = g(y)$ may be harder to solve, since it would involve an integral over f(y), though I would think you could just differentiate or Laplace transform to solve for f(y).

 

[pk415]

Thanks Mute

So then my integrating factor should be

$$e^{-xy}$$ right?

Then I have

$$V = e^{xy}[\int e^{-xy}e^x + f(y)]$$

$$V = e^{xy}[e^{1-y} \int e^x + f(y)]$$

$$V = e^{xy}[e^{1-y}e^x + f(y)]$$

$$V = e^{xy+x-y+1} + e^{xy}f(y)$$

$$U_y = e^{(x-1)y+x+1} + e^{xy}f(y)$$

And I was thinking I could bring the e^y into the generic function f(y) to get

$$U_y = e^{(x-1)y+x+1} + e^xf(y)$$

then integrate wrt y and get

$$U=\frac{1}{x-1}e^{(x-1)y+x+1} + e^xg(y) + h(x)$$

But when I put this U back into the original equation it doesn't work...

Does this make any sense?

 

[Mute]

The integrating factor is correct now, but you seem to be making the mistake $e^{ab} = e^ae^b$. The correct laws are $(e^a)^b = e^{ab}$ and $e^{a+b} = e^ae^b$.

So, you should get

$$e^{-xy}V = \left[\int dx~e^{-xy}e^x\right] + f(y)= \left[\int dx~e^{x(1-y)}\right] + f(y) = \frac{e^{-xy}e^x}{1-y} + f(y)$$