What is the solution to the Differential Equation e^(x+y)(dy/dx) = x with #y(0)=1?
1 Answer
y = ln (-xe^x - e^x + e+1)
Explanation:
We have:
e^(x+y)(dy/dx) = x withy(0)=1 ..... [A]
We can rearrange the DIfferential Equation [A] as follows:
e^xe^y dy/dx = x => e^y dy/dx = x e^(-x)
This is a First Order Separable Differential equation ad we now "seperate the variables" to get:
int \ e^y \ dy = int \ x e^(-x) \ dx ..... [B]
The LHS integral is a standard result, and for the RHS| integral we would need to apply Integration By Parts:
Let
{ (u,=x, => (du)/dx,=1), ((dv)/dx,=e^-x, => v,=-e^-x ) :}
Then plugging into the IBP formula:
int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx
gives us
int \ (x)(e^x) \ dx = (x)(-e^-x) - int \ (-e^-x)(1) \ dx
:. int \ xe^-x \ dx = -xe^-x - e^-x
Using this result, we can now integrate [B] to get the General Solution:
e^y = -xe^x - e^x + C
Using the initial condition
e^1 = -0e^0 - e^0 + C => C = e+1
Hence, the Particular Solution is:
e^y = -xe^x - e^x + e+1
And we can gain an explicit solution for [A] if we take Natural Logarithms:
ln (e^y) = ln (-xe^x - e^x + e+1 )
So that finally:
y = ln (-xe^x - e^x + e+1)
