# Another separable differential equation

• October 5th 2008, 02:29 PM
fastcarslaugh
Another separable differential equation
I understand that this requires integration by parts.

Find the function https://webwork.math.uga.edu/webwork...1f8b0486a1.png (for https://webwork.math.uga.edu/webwork...9363915f91.png ) which satisfies the separable differential equation with the initial condition https://webwork.math.uga.edu/webwork...cd9bbf64d1.png.

I've gotten to the point where

int y^2 dy = int (10+14x)/x dx

left side with y is straight forward, but the right side with x is ugly.

I made u = (10+14x) and dv = 1/x

du = 14 dx and v = ln(x)

after this i got something really ugly that i can't even type.

• October 5th 2008, 02:34 PM
Krizalid
Why make it so hard? Having $\frac{10+14x}x=\frac{10}x+14,$ and we're done.
• October 5th 2008, 04:27 PM
fastcarslaugh
yeah, i've tried that
if I still have to integrate (10/x + 14) by separation of variable
• October 5th 2008, 04:35 PM
skeeter
so integrate ...

$\int y^2 \, dy = \int \frac{10}{x} + 14 \, dx$

$\frac{y^3}{3} = 10\ln|x| + 14x + C$

$y^3 = 30\ln|x| + 42x + C_2$

$y = \sqrt[3]{30\ln|x| + 42x + C_2}$

now use your initial condition to determine $C_2$.