Find the solution of the given differential equation satisfying the indicated initial condition. y’ = (ln^2)y, y(2) = 4

Question

terenzreignz · Answer

Can't guarantee i can do this, but is it
$$\Large y\prime = \ln^2(y)$$

anonymous · Answer

yess my bad

terenzreignz · Answer

How do you usually do this, with series?

anonymous · Answer

@SithsAndGiggles

anonymous · Answer

$$\frac{dy}{dx}=\ln^2y$$
Now, do you mean $\ln^2y=\ln(\ln y)$, or $\ln^2y=(\ln y)^2$? There's some ambiguity with the notation.

anonymous · Answer

they way its written is (ln2)y

anonymous · Answer

attachment

anonymous · Answer

Well that explains it. $\ln2$ is a coefficient, approximately 0.69315.

The differential equation is separable:
$$\frac{dy}{dx}=\ln2 ~y\
\frac{1}{y}~dy=\ln2~dx$$
Integrate both sides:
$$\int\frac{1}{y}~dy=\int\ln2~dx\
\ln|y|=\ln2~x+C\
y=e^{\ln2~x+C}\
y=Ce^{\ln 2 ~x}\
y=C\left(e^{\ln 2}\right)^x\
y=C\cdot2^x$$
Now, given that $y(2)=4$, use that information to solve for $C$.

terenzreignz · Answer

Cr*p @SithsAndGiggles I thought it was $\large \ln^2(y)$  and went into a frenzy XD

anonymous · Answer

@terenzreignz, I thought so too. There's no way to integrate that, I don't think...