MHB Find a product solution to the following PDE

[Umar]

So I'm asked to use separation of variables to find a product solution to the given PDE:

(5y + 7)du/dx + (4x+3)du/dy = 0

Since it says to find a product solution, I used the form u(x,y) = XY and plugged that into the PDE. However, I am getting stuck because I'm not sure how exactly I should separate these two so that I can solve two different diff. equations. Your help is greatly aprpeciated!

 

[Opalg]

So I'm asked to use separation of variables to find a product solution to the given PDE:

(5y + 7)du/dx + (4x+3)du/dy = 0

Since it says to find a product solution, I used the form u(x,y) = XY and plugged that into the PDE. However, I am getting stuck because I'm not sure how exactly I should separate these two so that I can solve two different diff. equations. Your help is greatly appreciated!

When you plug $u = XY$ into the PDE, you get $(5y + 7)X'Y + (4x+3)XY' = 0$. Write that as $$\frac{(4x+3)X}{X'} = -\frac{(5y+7)Y}{Y'}.$$ The left side is a function of $x$ only, and the right side is a function of $y$ only. So both sides must be a constant, say $$\frac{(4x+3)X}{X'} = -\frac{(5y+7)Y}{Y'} = k.$$ That gives you your two equations, one for $X$ and one for $Y$.

 

[Umar]

When you plug $u = XY$ into the PDE, you get $(5y + 7)X'Y + (4x+3)XY' = 0$. Write that as $$\frac{(4x+3)X}{X'} = -\frac{(5y+7)Y}{Y'}.$$ The left side is a function of $x$ only, and the right side is a function of $y$ only. So both sides must be a constant, say $$\frac{(4x+3)X}{X'} = -\frac{(5y+7)Y}{Y'} = k.$$ That gives you your two equations, one for $X$ and one for $Y$.

Thanks for the reply! I actually got to that part, it's just that when I go to integrate the two functions, I would be integrating with respect to X no? So would I just treat the small x as a constant? Same with the y? What would I do with the constant k? Sorry for all the questions, but if you could show how you would solve one of the equations, I would really appreciate that. I'm also given two initial conditions for u, which I would apply after multiply X and Y to get the product solution..

 

[Opalg]

Thanks for the reply! I actually got to that part, it's just that when I go to integrate the two functions, I would be integrating with respect to X no? So would I just treat the small x as a constant? Same with the y? What would I do with the constant k? Sorry for all the questions, but if you could show how you would solve one of the equations, I would really appreciate that. I'm also given two initial conditions for u, which I would apply after multiply X and Y to get the product solution..

No, $X$ is meant to be a function of the variable $x$, and $Y$ is a function of $y$. Perhaps it would have been better to write $u(x,y) = f(x)g(y)$ instead of $u(x,y) = XY$. Then the equation would become $$\frac{(4x+3)f(x)}{f'(x)} = -\frac{(5y+7)g(y)}{g'(y)} = k.$$

 

[Umar]

No, $X$ is meant to be a function of the variable $x$, and $Y$ is a function of $y$. Perhaps it would have been better to write $u(x,y) = f(x)g(y)$ instead of $u(x,y) = XY$. Then the equation would become $$\frac{(4x+3)f(x)}{f'(x)} = -\frac{(5y+7)g(y)}{g'(y)} = k.$$

Okay so I got the following:

X = ((4x+3)*x^2)/(2*k) + C

Y = ((-5y+7)*y^2)/(2*k) + C

So the general solution would be the product of these two functions. I'm given two initial conditions:

u(0,0) = 6 and ux(0,0) = 9

With the first one, I get c = sqrt(6), but the second one doesn't even work out because plugging in x and y for 0's just eliminates everything. I feel like I'm making something wrong, and the value of K shouldn't even be there in the final answer.

 

[Opalg]

Okay so I got the following:

X = ((4x+3)*x^2)/(2*k) + C

Y = ((-5y+7)*y^2)/(2*k) + C

So the general solution would be the product of these two functions. I'm given two initial conditions:

u(0,0) = 6 and ux(0,0) = 9

With the first one, I get c = sqrt(6), but the second one doesn't even work out because plugging in x and y for 0's just eliminates everything. I feel like I'm making something wrong, and the value of K shouldn't even be there in the final answer.

Not sure how you got that. Write the equation $$\frac{(4x+3)f(x)}{f'(x)} = k$$ as $$\frac{f'(x)}{f(x)} = \frac{4x+3}k$$ and integrate both sides to get $$\ln(f(x)) = \frac{2x^2 + 3x}k + C$$, or $$f(x) = A\exp \bigl((2x^2 + 3x)/k\bigr)$$ (where $A,C$ denote a constant of integration). In a similar way, $$g(x) = B\exp \bigl((-\tfrac52y^2 - 7y)/k\bigr)$$. You can then use the initial conditions to find that $AB = 6$ and $k=2$.

I get the final answer to be $$u(x,y) = 6\exp\bigl(x^2 + \tfrac32x - \tfrac54y^2 - \tfrac72 \bigr).$$

 

[Umar]

Not sure how you got that. Write the equation $$\frac{(4x+3)f(x)}{f'(x)} = k$$ as $$\frac{f'(x)}{f(x)} = \frac{4x+3}k$$ and integrate both sides to get $$\ln(f(x)) = \frac{2x^2 + 3x}k + C$$, or $$f(x) = A\exp \bigl((2x^2 + 3x)/k\bigr)$$ (where $A,C$ denote a constant of integration). In a similar way, $$g(x) = B\exp \bigl((-\tfrac52y^2 - 7y)/k\bigr)$$. You can then use the initial conditions to find that $AB = 6$ and $k=2$.

I get the final answer to be $$u(x,y) = 6\exp\bigl(x^2 + \tfrac32x - \tfrac54y^2 - \tfrac72 \bigr).$$

Thank you so much for your time and help. I understand it now :)