Solve: dy/dx+ y = xy^3
.??

Question

Evelyn Agard · Accepted Answer

#y = pm 1/sqrt( x + 1/2 + C e^(2x))#

#dy/dx+ y = xy^3#

This is non-linear but we can make it linear using a Bernoulli substitution. Here we let:

Multiply both sides of the DE by #- 2/y^3#:

#underbrace(dy/dx * (- 2/y^3))_(= z')+ underbrace(y* (- 2/y^3))_(= - 2z) = underbrace(xy^3 * (- 2/y^3))_(= - 2x) #

For this, the integrating factor is: #exp(int (-2) dx) = e^(-2x)#

#e^(- 2x) (z' - 2z) [=(e^(- 2x) z)^'] = - 2xe^(- 2x)#

Integrating both sides, using IBP on RHS:

#e^(- 2x) z = int \ x \ d( e^(- 2x))#

#e^(- 2x) z = \ x \ e^(- 2x) - int dx \ e^(- 2x)#

#e^(- 2x) z = x e^(- 2x) + 1/2 \ e^(- 2x) + C#

# z = x + 1/2 + C e^(2x) qquad = 1/y^2#

#y = pm 1/sqrt( x + 1/2 + C e^(2x))#

Christopher Allers · Answer

undefined This is a first-order nonlinear ordinary differential equation. To solve it, we can use the method of separation of variables.

1. Rewrite the equation in the form $ \frac{dy}{dx} = xy^3 - y $.
2. Factor out $y$ on the right side to get $ \frac{dy}{dx} = y(x y^2 - 1) $.
3. Now, separate variables: $ \frac{1}{y} dy = (x y^2 - 1) dx $.
4. Integrate both sides. The integral of $ \frac{1}{y} dy $ is $ \ln|y| $, and the integral of $ (x y^2 - 1) dx $ requires a substitution.
5. Solve the resulting integrals to find the general solution.

This process will lead to the solution of the given differential equation.

Solve: dy/dx+ y = xy^3 .??