How to solve this Differential Equation?

Question

xy + y' = 160x

Answer 1

Since we have only a first derivate term, this is a linear first-order differential equation. This differential equation is usually in the form.

 

y' + p(x)y = g(x)

 

y' + xy = 160x

 

 

Next, we find the integrating factor, u(t).

 

u(t) = e^∫p(x)dt

       = e^∫xdx

       = e^{(1/2)x^2}

 

Multiply both sides of the equation by the integrating factor.

 

y'e^{(1/2)x^2} + xye^{(1/2)x^2} = 160xe^{(1/2)x^2}

 

 

You should end up with something like this.

 

d/dx[  ]= 160xe^{(1/2)x^2}

 

 

In this part, we must find the anti-derivative of the left side of the equation.  Apply the integration by parts method that you learned in your previous calculus classes.  Insert that anti-derivative into the bold brackets.

 

After that, integrate both sides of the equation.  This will undo the d/dx notation.  The left side will have a y term rather than a y' term.  Then use simple algebraic manipulation to solve for y.

 

Answer 2

To start solve for y'

 

y'=x(-y+160) now divide both sides by (-y+160)

 

y'/(-y+160)=x now integrate both sides wrt x

 

∫dy/dx/(-y+160)dx=∫xdx

 

-ln(-y+160)=1/2x²+c where c i a constant now solve for y and you have the answer

 

Hope this helps

 

Jim