Toronto Math Forum

a) Find the general solution of  $y’’+2y’+17y=40e^x + 130 \sin(4x)$

First, for the complimentary solution, consider the homogeneous equation:
$y’’+2y’+17y=0$

and then, the characteristic equation is: $r^2+2r+17=0$,
$r \ = \ \frac{-2 \pm \sqrt{4-4 \times 17}}{2} \ = \ \frac{-2 \pm 8i}{2} \ = \ -1 \pm 4i$

These roots are a pair of complex conjugates in the form of $ \lambda \pm i \mu$,
so the differential equation has a general solution in the form of $y(x)=C_{1} e^{\lambda x} \cos(\mu x)+C_{2} e^{\lambda x} \sin(\mu x)$.

In this case, we have: $y_{c}(x)= e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))$

Then, for the particular solution, by the method of undetermined coefficients,
suppose $y_{p}=A \cdot e^x$ satisfies the equation: $y’’+2y’+17y=40e^x$

Since, $y_{p}=A \cdot e^x$
$y’_{p}=A \cdot e^x$
$y’’_{p}=A \cdot e^x$

plug into the equation: $A \cdot e^x + 2A \cdot e^x + 17 A \cdot e^x = 20A \cdot e^x = 40e^x$
so, $A=2$
$\therefore y_{p}=2e^x$

For another particular solution, suppose $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$ satisfies the equation: $y’’+2y’+17y=130 \cdot \sin(4x)$

Since, $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$
$y’_{p}=-4B \cdot \sin(4x) + 4C \cdot \cos(4x)$
$y’’_{p}=-16B \cdot \cos(4x) -16 C \cdot \sin(4x)$

plug into the equation:  $-16B \cdot \cos(4x) -16 C \cdot \sin(4x)-8B \cdot \sin(4x) + 8C \cdot \cos(4x) + 17B \cdot \cos(4x) + 17C \cdot \sin(4x) \\ =(B+8C) \cdot \cos(4x) + (C-8B) \cdot \sin(4x) \\ =130 \sin(4x)$

Then, $
\begin{cases}
C-8B=130 \\
B+8C=0
\end{cases}$

=> $\begin{cases}
C=2 \\
B=-16
\end{cases}$

$\therefore y_{p}=-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

So, the general solution is: $y(x)=e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$


b) when $y(0)=0, y’(0)=0$, and we have
$y’(x)=-e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+e^{-x} (-4C_{1} \cdot \sin(4x) +4C_{2} \cdot \cos(4x))+2e^x+64 \cdot \sin(4x) + 8 \cdot \cos(4x)$,

plug $y(0)=0$ into $y(x)$ equation, we get: $0=C_{1}+2-16$ => $C_{1}=14$,

then, plug $y’(0)=0$ into $y’(x)$ equation, we have:
$-C_{1}+4C_{2}+2+8=0 \\
-14+4C_{2}+10=0 \\
C_{2} = 1$

$\therefore y(x)=e^{-x} (-14\cdot \cos(4x) + \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

$$1 + (\frac{x}{y}-\sin(y))y’ = 0$$
Let $M=1, N=\frac{x}{y}-\sin(y)$
Then, we can get: $M_{y}=0, N_{x}=\frac{1}{y}$
Define $R=\frac{M_{y}-N_{x}}{M}=\frac{-\frac{1}{y}}{1}=-\frac{1}{y}$,
=> The integrating factor should be: $\mu (x,y) = e^{-\int R dy} = e^{ln|y|} = y$

Multiple both sides by y, we can define the new $M = y, N = x-\sin(y)y$,

Since it’s exact now, there exists a function $\phi (x,y)$, such that: $\phi_{x}(x,y) = M(x,y)\ and\ \phi_{y}(x,y) = N(x,y)$
Integrating with respect to x:
$\phi(x,y) = \int M(x,y) \ dx \\ i.e. \phi(x,y) = \int y\ dx \\ \phi(x,y) = xy + h(y)$
Then, we have: $\phi_{y} =  x + h’(y) = N = x-y\sin(y)$, i.e. $h’(y)=-y\sin(y)$
=> $h(y) = -\int y \sin(y) \ dy$

Using integrating by parts,  $u = y, \ v = -\cos(y),\\du = 1, \ dv = \sin(y) $

=> $h(y) = -(-y \cos(y)- \int(-\cos(y))\ dy) = y \cos(y) - \sin(y)$

So, the solution of the given equation is: $\phi(x,y) = xy + y \cos(y) - \sin(y) = C$, where C is the constant