Finding a partial solution to the differential equation Given, $$\frac{dy}{dx} = yx \sin(x), \text{ when }y(0) = 1/4.$$
What I did is first, separate the equation
 $$ \frac{dy}{y} = x \sin (x) dx. $$
Second, integrate by parts
 $$ \ln |y| + C1 = -x \cos (x) + \sin (x) + C2 \Rightarrow C3 = C2-C1.$$
Third, left |y| alone
 $$|y| = e^{-x \cos (x) }+ e^{\sin(x)} + e^{C3}.$$
Fourth, inserted the values $$y(0) = 1/4
 \Rightarrow 1/4 = 1 + 1 + e^{C3}
 \Rightarrow -1.75 = e^{C3}
 \Rightarrow \ln (-1.75) = \ln e^{C3}.
$$
Here, since negative doesn't work inside ln ( ) I think there is something wrong with my solution. Can anyone help me with how I should approach this question?
Thank you!
 A: There's nothing wrong with the method, you just messed up you exponentiation. When the power is a sum, you end up with a product, not a sum!
A: you can separate as  $$\ln(4y) = \ln y - \ln (1/4)=\int_{1/4}^y \frac{dy}{y} = \int_0^x x \sin x \, dx = -x\cos x\big|_0^x + \int_0^x \cos x \, dx = - x\cos x+\sin x$$
so $$y = \frac 1 4e^{\sin x - x\cos x} $$
A: As well as separating the variables, you can use the integrating factor if you like. Write
\begin{equation*}
\frac{dy}{dx}-xy\sin(x)=0.
\end{equation*}
Then we have 
\begin{equation*}
e^{-\int x\sin(x)dx}=e^{x\cos(x)-\sin(x)}.
\end{equation*}
Multiplying each term by this and using the reverse product rule gives
\begin{equation*}
\frac{d}{dx}(e^{x\cos(x)-\sin(x)}y)=0.
\end{equation*}
Integrate both sides with respect to $x$ and apply the initial condition. Does that help?
A: $$
y'=yx\sin x\quad\Longrightarrow\quad
\exp\left(-\int_0^x t\sin t\,dt\right)(y'-yx\sin x)=0\quad\Longrightarrow\quad
\bigg(y(x)\exp\Big(-\int_0^x t\sin t\,dt\Big)\bigg)'=0\quad\Longrightarrow\quad
y(x)\exp\Big(-\int_0^x t\sin t\,dt\Big)=c,
$$
for some constant $c$. In fact, $y(0)=1/4$, implies that $c=4$. Thus
$$
y(x)=\frac{1}{4}\exp\Big(\int_0^x t\sin t\,dt\Big)=
\frac{1}{4}\exp\Big(-x\cos x+\sin x\Big).
$$
