Solve the initial value problem

[Lovely918]

Well I have an initial value problem:
y+ dy/dx -xy^3=0, y(0)=2

So I've been working on this and below is my attempt (please tell me if im doing it right). Also I have no idea what to do with the condition of y(0)=2
dy/dx +y=xy^3
Where IF is of the form
dy/dx+Py=Q so this means p=1, Q=xy^3
so
e^ integral (Pdx) =e^integral(1dx) = e^x
solution is given by
y= Integral (Q(I.F.) dx
=integral (x(y^3)*e^x dx
y^3 Integral (x(e^x) dx)
y/y^3 =x(e^x) -e^x +c
So what do I do now? How do I solve the rest of it with the condition? Thank you for all of your time.

 

[mmm4444bot]

Where IF is of the form
dy/dx + Py = Q

The equation above is not the integrating factor (IF).

The IF would be e^integral(P dx).

this means p=1, Q=xy^3

I have not used this method for a long time, but I'm thinking that it requires P and Q to each be either a constant or a function of x only. (Your Q is a function of x and y.)

Have you learned any other methods?

 

[HallsofIvy]

Well I have an initial value problem:
y+ dy/dx -xy^3=0, y(0)=2

Theorectically, every first order differential equation has an integrating factor but the method you use to find the integrating factor works only for linear equations and, because of the \(\displaystyle y^3\), this is not linear.

Though it is not exactly an "integrating factor" for this equation, you can make use of the fact that \(\displaystyle d(e^xy)/dx= e^xdy/dx+ e^xy= e^x(dy/dx+ y)\), writing the equation as \(\displaystyle d(e^xy)/dx= xe^xy^3\). Now, let \(\displaystyle u= e^xy\) so that \(\displaystyle y= e^{-x}u\) and \(\displaystyle y^3= e^{-3x}u^3\) so the equation becomes \(\displaystyle \frac{du}{dx}= xe^{-3x}u^3\) which is separable:
\(\displaystyle \frac{du}{u^3}= xe^{-3x}dx\).

Integrate both sides of that.

So I've been working on this and below is my attempt (please tell me if im doing it right). Also I have no idea what to do with the condition of y(0)=2
dy/dx +y=xy^3
Where IF is of the form
dy/dx+Py=Q so this means p=1, Q=xy^3
so
e^ integral (Pdx) =e^integral(1dx) = e^x
solution is given by
y= Integral (Q(I.F.) dx
=integral (x(y^3)*e^x dx
y^3 Integral (x(e^x) dx)
y/y^3 =x(e^x) -e^x +c
So what do I do now? How do I solve the rest of it with the condition? Thank you for all of your time.

Because the equation was not linear, no, this is not correct. If it were, you "solve the rest of it with the condition", y(0)= 2, by replacing x with 0, y with 2, and solving the resulting equation for c.

 

[Lovely918]

Thank you!

Thank you for all of your help!

Theorectically, every first order differential equation has an integrating factor but the method you use to find the integrating factor works only for linear equations and, because of the \(\displaystyle y^3\), this is not linear.

Though it is not exactly an "integrating factor" for this equation, you can make use of the fact that \(\displaystyle d(e^xy)/dx= e^xdy/dx+ e^xy= e^x(dy/dx+ y)\), writing the equation as \(\displaystyle d(e^xy)/dt= xe^xy^3\). Now, let \(\displaystyle u= e^xy\) so that \(\displaystyle y= e^{-x}u\) and \(\displaystyle y^3= e^{-3x}u^3\) so the equation becomes \(\displaystyle \frac{du}{dx}= xe^{-3x}u^3\) which is separable:
\(\displaystyle \frac{du}{u^3}= xe^{-3x}dx\).

Integrate both sides of that.


Because the equation was not linear, no, this is not correct. If it were, you "solve the rest of it with the condition", y(0)= 2, by replacing x with 0, y with 2, and solving the resulting equation for c.