Home > Calculus > Solving Separable Differential Equations

What is the general solution of the differential equation # y'=y/x+xe^x #?

Answer 1

# y = xe^x + Cx #

We are trying to solve the First Order ODE:

# y'=y/x+xe^x #

When we have a First Order Linear non-homogeneous Ordinary Differential Equation of the following form, we can use an integrating factor;

# dy/dx + P(x)y=Q(x) #

So, in standard form, rewrite the equations as follows:

# dy/dx - y/x = xe^x ..... [1] #

Next, the following gives the integrating factor:

# I = e^(int P(x) dx) # # \ \ = exp(int \ -1/x \ dx) # # \ \ = exp( -lnx ) # # \ \ = 1/x #

And if we multiply the DE [1] by this Integrating Factor, #I#, we will have a perfect product differential;

# 1/xdy/dx - y/x^2 = e^x #

# :. d/dx( y/x ) = e^x #

which we can integrate directly to obtain:

# y/x = int \ e^x + C #

# :. y/x = e^x + C #

# :. y = xe^x + Cx #

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Christopher Allers

The general solution of the given differential equation (y' = \frac{y}{x} + xe^x) is:

[y = x(e^x + C)]

where (C) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.