How do you solve the fallowing ode?
u'=u^2
What I did was:
$ \frac{du}{dt}=u^2 \rightarrow du=u^2dt\rightarrow\int du=\int u^2dt\rightarrow u=u^2(t+c)\rightarrow u=\frac{1}{t+c} $
but the correct answer is $ u=\frac{1}{-t+c} $ Where was I wrong?
How do you solve the fallowing ode?
u'=u^2
What I did was:
$ \frac{du}{dt}=u^2 \rightarrow du=u^2dt\rightarrow\int du=\int u^2dt\rightarrow u=u^2(t+c)\rightarrow u=\frac{1}{t+c} $
but the correct answer is $ u=\frac{1}{-t+c} $ Where was I wrong?
Here is the correct way to do it:
First, divide by $u^2$ to get
u'/u^2=1.
Then integrate both sides to get
$-1/u=t+d$.
Finally, rearranging and letting $c=-d$ yields
$u=\frac{1}{c-t}$
as desired.
Use the method of separation of variables:
$\frac{du}{dt}=u^2\Longrightarrow u^{-2}du=dt,$ $\int u^{-2}du=\int dt\Longrightarrow -u^{-1}=t+c_1\Longrightarrow u=\frac{1}{c_1-t}.$
Remember that $c_1$ can 'absorb' the negative sign.