- #1
sourlemon
- 53
- 1
[SOLVED] Seperable Equation
1. Instruction: Solve the equation.
2. Equations:
dy/dx = g(x)p(y)
h(y) = 1/p(y)
h(y)dy = g(x)dx
[tex]\int[/tex]h(y)dy = [tex]\int[/tex]g(x)dx
H(y) = G(x) + C
3. http://img354.imageshack.us/img354/6475/mathdl0.jpg
I tried to do it on the right side, but...I got stuck there. If I add to the right, then I would be left with -C + e[tex]^{-y}[/tex] = ex + -ye[tex]^{-y}[/tex]. Can I say that -C = C? But what about e[tex]^{-y}[/tex] . Did I inegrate it right?
Thank you in advance.
1. Instruction: Solve the equation.
2. Equations:
dy/dx = g(x)p(y)
h(y) = 1/p(y)
h(y)dy = g(x)dx
[tex]\int[/tex]h(y)dy = [tex]\int[/tex]g(x)dx
H(y) = G(x) + C
3. http://img354.imageshack.us/img354/6475/mathdl0.jpg
I tried to do it on the right side, but...I got stuck there. If I add to the right, then I would be left with -C + e[tex]^{-y}[/tex] = ex + -ye[tex]^{-y}[/tex]. Can I say that -C = C? But what about e[tex]^{-y}[/tex] . Did I inegrate it right?
Thank you in advance.
Last edited by a moderator: