mathstresser
Junior Member
- Joined
- Jan 28, 2006
- Messages
- 134
Find the general solution of the indicated differential equation.
\(\displaystyle \L\\ xy'\, -\, y\, =\, 2x^2 y\)
*That is xy-prime - y =...
So, I get...
\(\displaystyle \L\\ x\, \frac{dy}{dx}\, =\, 2x^2 y\,dx\, +\, y\, dx\)
\(\displaystyle \L\\ x\,dy\,=\, dx(2x^2 y\, +\, y)\)
\(\displaystyle \L\\ x\,dy\, =\, dx(y)(1\,+\,2x^2)\)
\(\displaystyle \L\\ \frac{dy}{y}\,=\, \left(\,\frac{1\,+\,2x^2}{x}\,\right)\,dx\)
The last part is where I have problems. I can do the first integral, but not the second.
ln y = ???
\(\displaystyle \L\\ xy'\, -\, y\, =\, 2x^2 y\)
*That is xy-prime - y =...
So, I get...
\(\displaystyle \L\\ x\, \frac{dy}{dx}\, =\, 2x^2 y\,dx\, +\, y\, dx\)
\(\displaystyle \L\\ x\,dy\,=\, dx(2x^2 y\, +\, y)\)
\(\displaystyle \L\\ x\,dy\, =\, dx(y)(1\,+\,2x^2)\)
\(\displaystyle \L\\ \frac{dy}{y}\,=\, \left(\,\frac{1\,+\,2x^2}{x}\,\right)\,dx\)
The last part is where I have problems. I can do the first integral, but not the second.
ln y = ???