What is the solution to the Differential Equation #e^(x+y)(dy/dx) = x# with #y(0)=1?

Question

Isaiah Allen · Accepted Answer

# y = ln (-xe^x - e^x + e+1) #

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 #y(0)=1# we have: # 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) #

Isaiah Allen · Answer

undefined To solve the differential equation $ e^{x+y} \frac{dy}{dx} = x $ with the initial condition $ y(0) = 1 $, we can use the method of separation of variables.

1. Rearrange the equation to isolate $ \frac{dy}{dx} $:
   $ \frac{dy}{dx} = \frac{x}{e^{x+y}} $

2. Separate variables:
   $ e^{y} dy = x e^{-x} dx $

3. Integrate both sides:
   $ \int e^{y} dy = \int x e^{-x} dx $
   Integrating $ e^{y} $ with respect to $ y $ gives $ e^{y} $.
   Integrating $ x e^{-x} $ with respect to $ x $ requires integration by parts:
   Let $ u = x $ and $ dv = e^{-x} dx $, then $ du = dx $ and $ v = -e^{-x} $.
   Using integration by parts, we get:
   $ \int x e^{-x} dx = -xe^{-x} - \int -e^{-x} dx = -xe^{-x} + e^{-x} + C_1 $,
   where $ C_1 $ is the constant of integration.

4. Substitute back and simplify:
   $ e^{y} = -xe^{-x} + e^{-x} + C_1 $
   $ y = \ln(-xe^{-x} + e^{-x} + C_1) $

5. Apply the initial condition $ y(0) = 1 $ to find $ C_1 $:
   $ 1 = \ln(-0e^{0} + e^{0} + C_1) $
   $ 1 = \ln(1 + C_1) $
   $ e = 1 + C_1 $
   $ C_1 = e - 1 $

6. Substitute $ C_1 $ back into the solution:
   $ y = \ln(-xe^{-x} + e^{-x} + e - 1) $