# 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# with#y(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) #