# Math Help - [SOLVED] initial value differential equation

1. ## [SOLVED] initial value differential equation

(x)dy/dx + 2y = 4x^2
y(1) = 2

Solve the initial value problem. Once I get the variable separated I know how to solve the problem. I just can't figure out how to get the x out of the left side. Please help!!

2. Hello, mathprincess24!

$x\frac{dy}{dx} + 2y \:=\: 4x^2,\;\;y(1) = 2$

Solve the initial value problem.
We can't separate the variables . . . is that the only method you know?

Divide by $x\!:\;\;\frac{dy}{dx} + \frac{2}{x}\,y \;=\;4x$

Integrating factor: . $I \;=\;e^{\int\frac{2}{x}dx} \;=\;e^{2\ln x} \:=\:e^{\ln x^2} \:=\:x^2$

Multiply by $I\!:\;\;x^2\frac{dy}{dx} + 2xy \:=\:4x^3$

And we have: . $\frac{d}{dx}\left(x^2y\right) \;=\;4x^3$

Integrate: . $x^2y \;=\;x^4 + C\quad\Rightarrow\quad y \;=\;x^2 + \frac{C}{x^2}$

Since $y(1) = 2$, we have: . $2 \:=\:1^2 + \frac{C}{1^2} \quad\Rightarrow\quad C \,=\,1$

Therefore: . $y \;=\;x^2 + \frac{1}{x^2}$

3. Originally Posted by mathprincess24
(x)dy/dx + 2y = 4x^2
y(1) = 2

Solve the initial value problem. Once I get the variable separated I know how to solve the problem. I just can't figure out how to get the x out of the left side. Please help!!
You can't. That's not a "separable" equation. It is, however, a linear equation. If you divide by x, you have dy/dx+ 2y/x= 4x. Now you can find an "integrating factor", a function m(x), such that multiplying by it makes the left side a "complete" derivative: d(my)/dx= m dy/dx + (dm/dx)y which much be equal to m dy/dx+ (2/x)y. That means we must have dm/dx= m(2/x) which IS a separable equation: dm/m= 2dx/x which integrates to ln(m)= 2 ln(x)= ln(x^2) or y= x^2 (you can ignore the "constant of integration" because we just want one possible solution). If you multiply the entire equation by x^2 we get x^2 dy/dx+ 2xy= d(x^2y)/dx= 4x^3 which we can now write as d(x^2y)= 4x^4 dx and integrate to get x^2y= x^4+ C. Set x= 1, y= 2 to find C.

Once again Soroban beat me! And now I have been beaten by AIR by a nose!

4. thank you so much. that really helps. we learned the integrating factor. i just have a hard time sometimes recognizing when to use it if the problem isnt in the exact general form. thanks again!!