$$\frac{\partial^2z}{\partial x^2}=10x^3y+12x^2y^2$$ I need to find the particular solution $z(x,y)$ subject to the boundary conditions $$z=y \space and \space \frac{\partial z}{\partial x}=y^2 \space when \space x=1$$

I integrated with respect to x and found: $10y\frac{x^4}{4}+4x^3y^2+f(y)$ then i integrated again this with respect to x and found:$2x^5y+x^4y^2+f(y)x+g(y)$

How do i continue from here to find the particular solution?


From integrating twice, we find that


Next, we apply the conditions at $x=1$, $z(1,y)=y$ and $z_x(1,y)=y^2$. Then, we find that

$$\frac12y+y^2+f(y)+g(y)=y \tag 1$$


$$\frac52y+4y^2+f(y)=y^2\tag 2$$

From $(2)$ we can determine $f(y)$ and using this form for $f(y)$ and substituting it into $(1)$ we can determine $g$.

Can you finish this now?

  • $\begingroup$ Pleased to see that this was useful! Since you are new to the site, I wanted to let you know that once you have enough reputation points, you can up vote answers also. -Mark $\endgroup$ – Mark Viola Apr 1 '17 at 5:28

Check your work, you integrated twice with respect to $x$ and should have: $$z=\frac12x^5y+x^4y^2+xf(y)+g(y)$$

You need a particular solution such that: $$z|_{x=1}=\frac12y+y^2+f(y)+g(y)=y$$ $$\frac{\partial z}{\partial y}|_{x=1}=\frac52y+4y^2+f(y)=y^2$$

Solving the second equation gives $f$, then the first gives $g$. They are: $$f(y)=-3y^2-\frac52y$$ $$g(y)=2y^2+2y$$


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