Toronto Math Forum

Find the solution of the given initial value problem.
\begin{align*}
y''+y&=0, \\
y(\pi /3)&=2, \\
y'(\pi /3)&=-4
\end{align*}
Solution:
First we solve the characteristic equation:
\begin{align*}
r^2+1&=0\\
r&=\pm i
\end{align*}
so here $\lambda = 0, \mu = 1$, the general solution for the homogeneous equation is
\begin{align*}
y_1=c_1\cos(t)+c_2\sin(t)
\end{align*}
We can check the Wronskian is $1\neq 0$.
Differentiate y, we get
\begin{align*}
y'=-c_1\sin(t)+c_2\cos(t)
\end{align*}
Now substitute the initial values:
\begin{align*}
    y(\pi /3)&=2 \\
    c_1\cos(\pi /3) +c_2\sin(\pi/3) &=2\\
    \frac{1}{2}c_1+\frac{\sqrt{3}}{2}c_2 &=2\\
    y'(\pi /3) &=-4\\
    -c_1\sin(\pi/3)+c_2\cos(\pi/3)&=-4\\
    -\frac{\sqrt{3}}{2}c_1+\frac{1}{2}c_2 &=-4
\end{align*}
Solve for $c_1, c_2$, we get
\begin{align*}
c_1 = 1+2\sqrt{3}, c_2 = \sqrt{3}-2
\end{align*}
Then the solution of the initial value problem is
\begin{align*}
y = (1+2\sqrt{3})\cos(t) + (\sqrt{3}-2)\sin(t)
\end{align*}

Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
\begin{align*}
(\frac{\sin(y)}{y}- 2e^{-x}\sin(x))+(\frac{\cos(y)+2e^{-x}\cos(x)}{y})y' = 0, \mu(x,y) = ye^{x}.
\end{align*}
We first need to check if $M_y=N_x$ to decide if the equation is exact.
Notice
\begin{align*}
    M &= \frac{\sin(y)}{y}- 2e^{-x}\sin{(x)} \\
    N &=\frac{\cos(y)+2e^{-x}\cos(x)}{y}
\end{align*}
Then
\begin{align*}
    M_y &= \frac{\cos(y)y-\sin(y)}{y^2} \\
    N_x &= (\frac{\cos(y)}{y}+\frac{2e^{-x}\cos(x)}{y})_x \\
    &= \frac{-2e^{-2}\cos(x) - 2e^{-x}\sin(x)}{y}
\end{align*}
We showed the given equation is not exact since $M_y \neq N_x$.
Then we try to multiply the equation by the integrating factor $\mu(x,y) = ye^x$. Then we get
\begin{align*}
(\sin(y)e^{x}- 2y\sin{(x)})+(e^{x}\cos(y)+2\cos(x))y' = 0
\end{align*}
Now we check $M_y$ and $N_x$ of the new equation.
Notice
\begin{align*}
    M &= \sin(y)e^{x}- 2y\sin{(x)}\\
    N &= e^{x}\cos(y)+2\cos(x)\\
\end{align*}
Then
\begin{align*}
    M_y &= e^x \cos(y)-2\sin(x) \\
    N_x &= e^x\cos(y)-2\sin(x)
\end{align*}
We verified that $M_y = N_x$, so the new equation is exact.
Now we integrate $M$ to solve for the general equation.
\begin{align*}
\psi = \int (\sin(y)e^{x}- 2y\sin{(x)})dx = \sin(y)e^{x}+ 2y\cos{(x)}+h(y)
\end{align*}
where $h(y)$ is a function of just y and we can find it by partial differentiate $\psi$ with respect to $y$. So
\begin{align*}
\psi_y = e^x\cos(y)+2\cos(x) + h'(y) = N(x,y) = e^{x}\cos(y)+2\cos(x)
\end{align*}
so we get $h'(y)=o$ and $h(y)=c, c$ is a constant. Then
\begin{align*}
\psi = \sin(y)e^{x}+ 2y\cos{(x)} +c
\end{align*}
Therefore we get the general solution
\begin{align*}
\sin(y)e^{x}+ 2y\cos{(x)} = C
\end{align*}