- #1
jacobrhcp
- 169
- 0
[SOLVED] differential equation
[tex]\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0 [/tex]
I have made two attempts, both using the same substitution, where I think I made an error.
1.
[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=u^{2}, [/tex]
[tex] -\frac{1}{u}=x+c_{0}, [/tex]
[tex] y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0, [/tex]
[tex] y=-x [/tex]
checking the solution gives -1=0, which is false.
2.
[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=1, [/tex]
[tex] (x+y)=x+c_{0}, [/tex]
[tex] y=c_{0}=0 [/tex],
which gives 0=x^{2} after putting it back in the differential equation.I'm sure I did something fundamentally wrong, and I don't know why I would be allowed to make these substitutions. I did them because every other way of solving this led to something I didn't believe. Could anyone enlighten me on the idea of substitution?
Homework Statement
[tex]\frac{\partial y}{\partial x}=(x+y)^{2}, y(0)=0 [/tex]
The Attempt at a Solution
I have made two attempts, both using the same substitution, where I think I made an error.
1.
[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=u^{2}, [/tex]
[tex] -\frac{1}{u}=x+c_{0}, [/tex]
[tex] y=\frac{x^{2}+c_{0}+1}{x}, y(0)=0, [/tex]
[tex] y=-x [/tex]
checking the solution gives -1=0, which is false.
2.
[tex] u=x+y, \partial u = \partial y, [/tex]
[tex] \frac{\partial u}{\partial x}=1, [/tex]
[tex] (x+y)=x+c_{0}, [/tex]
[tex] y=c_{0}=0 [/tex],
which gives 0=x^{2} after putting it back in the differential equation.I'm sure I did something fundamentally wrong, and I don't know why I would be allowed to make these substitutions. I did them because every other way of solving this led to something I didn't believe. Could anyone enlighten me on the idea of substitution?
Last edited: