Separable Differential Equation with Initial Value?

Question

#x^2(dy/dx)=y-xy#

Initial condition : y(-1)=-1

This is what I separated it to

#(dy/y)=(1-x)/(x^(2))dx#

After integration, I have:

#ln(y)=(-xln(x)-1)/(x)#

I know that I will have to introduce "e," but I am getting tripped up here.

Any help would be appreciated.

Thanks!

Evelyn Agard · Accepted Answer

# |y| = (e^(-1/x-1))/(|x|) #

We have: # x^2 \ dy/dx = y-xy # with Initial condition #y(-1)=-1# which we could type as: # x^2 \ dy/dx = y(1-x) => 1/y \ dy/dx = (1-x)/x^2 # Since the ODE can be divided, as shown, we can "separate the variables" to obtain: # int \ 1/y \ dy = int \ 1/x^2 - 1/x \ dx # Since both integrals pertain to standard functions, we can integrate them directly to obtain: # ln |y| = -1/x - ln |x| + A # Using the initial condition #y(-1)=-1# we have: # ln |-1| = -1/(-1) - ln |-1| + A => A = -1 # Consequently, we have: # ln |y| = -1/x - ln |x| + 1 # Noting that we can write #lne=1# we get: # ln |y| = -1/x - lne - ln |x| # # :. ln |y| + ln |x| + lne = -1/x # # :. ln |xye| = -1/x # # :. |xye| = e^(-1/x) # # :. |y| = (e^(-1/x))/(e|x|) # # :. |y| = (e^(-1/x-1))/(|x|) #

Evelyn Agard · Answer

undefined A separable differential equation is one that can be expressed in the form $ \frac{dy}{dx} = f(x)g(y) $, where $ f(x) $ is a function of $ x $ and $ g(y) $ is a function of $ y $. To solve a separable differential equation with an initial value, you typically follow these steps:

1. Separate the variables: Rewrite the equation so that all terms involving $ y $ are on one side and all terms involving $ x $ are on the other side.

2. Integrate both sides: Integrate both sides of the equation with respect to their respective variables.

3. Solve for $ y $: After integrating, you'll have an equation involving $ y $ and $ x $. Solve this equation for $ y $ to find the general solution.

4. Apply the initial condition: Use the initial value given to find the particular solution by substituting the initial value into the general solution and solving for any constants.

5. Check for any restrictions: Ensure that the solution obtained is valid for all values of $ x $ within the domain of the original problem.

This process yields the solution to the separable differential equation with the given initial value.

Separable Differential Equation with Initial Value?

#x^2(dy/dx)=y-xy# Initial condition : y(-1)=-1 This is what I separated it to #(dy/y)=(1-x)/(x^(2))dx# After integration, I have: #ln(y)=(-xln(x)-1)/(x)# I know that I will have to introduce "e," but I am getting tripped up here. Any help would be appreciated. Thanks!

#x^2(dy/dx)=y-xy#

Initial condition : y(-1)=-1

This is what I separated it to

#(dy/y)=(1-x)/(x^(2))dx#

After integration, I have:

#ln(y)=(-xln(x)-1)/(x)#

I know that I will have to introduce "e," but I am getting tripped up here.

Any help would be appreciated.

Thanks!