(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!!

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- Feb 9th 2009, 10:58 AMmathprincess24[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!! - Feb 9th 2009, 11:54 AMSoroban
Hello, mathprincess24!

Quote:

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

Solve the initial value problem.

**can't**separate the variables . . . is that the only method you know?

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

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

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

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

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

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

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

- Feb 9th 2009, 12:01 PMHallsofIvy
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! - Feb 9th 2009, 03:52 PMmathprincess24
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!!