Question

Solve the initial-value problem
y'+6xy=0;y(pi)=5

Answer 1

\[\frac{dy}{dx}=-6xy \rightarrow dy*\frac{1}{y}=-6x dx\]
\[\int\limits y^{-1}dy=-6 \int\limits x dx \implies \ln|y|=-3x^2+C\]
\[\huge y=e^{-3x^2+c} \implies y=Ce^{-3x^2}\]
\[\huge 5=Ce^{-3(\pi^2)} \implies 5e^{3\pi^2}=C\]

Answer 2

I got that as well but the book says the answer is 5e^-3(x^2-pi^2)

Answer 3

Oh, well then you do this:
\[\huge y=Ce^{-3x^2}; C=5e^{3\pi^2} \implies y=5e^{3\pi^2}e^{-3x^2}\]
\[\huge =5e^{-3(x^2-\pi^2)}\]
Using the property that:
\[(e^a)(e^b)=e^{a+b}\]