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

1 Answer
Jul 5, 2018

Answer:

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

Explanation:

#dy/dx+ y = xy^3#

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

  • #z = 1/y^2 qquad qquad z' = - 2/y^3 y'#

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))#