Solving Differential Equations: Steps & Solutions

[Mark44]

Ok, for problem 6, I got dy/dx + 2y/x = 1/x by re-arranging and dividing by x. Since 2y/x is the coefficient then the integrating factor is the same as before x^2.

multiply all sides by x^2 I get, dy/dx(x^2)+ 2xy = x, next we integrate so

x^3/3+c = x^2/2 but how do I get the answer I provided in the beginning with this result?

If you integrate y'x² + 2xy, you don't get x³, either with or without a constant.

Let's look at a different example, and work it through. Hopefully you will follow what's going on and be able to apply it to your problems.


Problem: Solve y' -4y = 12

This DE is linear, so we can solve it by using an integrating factor. In this case, the integrating factor is
e^{\int -4 dx} = e^{-4x}
Multiply both sides of the equation by the integrating factor, getting
y'e^-4x -4ye^-4x = 12e^-4x

The two terms on the left side are the derivative of ye^-4x, which you can check by using the product rule. In other words, d/dx(ye^-4x) = y'e^-4x -4ye^-4x.
The whole purpose of the business of an integrating factor is to multiply y' +p(x)y by whatever it takes so that it becomes the derivative of something.

We're now ready to integrate both sides.

\int (y'e^{-4x} -4ye^{-4x})dx = \int 12e^{-4x} dx

The left side can be simplified, by replacing the integrand with something equal to it, and integrating the right side.
\int d/dx(ye^{-4x})dx = -3e^{-4x} + C

Then, since the integral of the derivative of something is just the something, we have
ye^-4x = -3e^-4x + C

If we solve for y by multiplying both sides by e^4x, we get
y = -3 + Ce^4x, the solution to our DE. It's easy to check that this is the solution.

 

[madahmad1]

thx a lot mark but one last favor, can u please show me how u checked using the product rule?

 

[Mark44]

Sure.
d/dx(ye^-4x) = y'e^-4x + y*(-4 e^-4x) = y'e^-4x - 4xye^-4x.

So y'e^-4x - 4xye^-4x is the derivative, with respect to x, of ye^-4x