Question

Solve: $\left( {{D}^{2}}+4D+13 \right)y=\cos 3x$.

Answer 1

The given second order differential equation has a function of x. So, this becomes a particular integral. As the function is trigonometrical, we assume the integral function as $y\left( x \right)=A\cos 3x+B\sin 3x$. We find the values of ${{D}^{2}}=\dfrac{{{d}^{2}}}{d{{x}^{2}}},D=\dfrac{d}{dx}$ and place them in the main equation. We get two equations of two unknowns A and B. we solve them to find the values of A and B.

Complete step by step answer:
We have been given a second order differential equation with constant coefficient.
$\left( {{D}^{2}}+4D+13 \right)y=\cos 3x$. Here ${{D}^{2}}=\dfrac{{{d}^{2}}}{d{{x}^{2}}},D=\dfrac{d}{dx}$.
In this type of equation, we get the characteristics equation by taking the differential form of the equation.
We assume the solution of the differential equation. As the function of x is $f\left( x \right)=\cos 3x$. We take the PI as $y\left( x \right)=A\cos 3x+B\sin 3x$.
Here PI describes a particular integral which is the solution of the differential equation $\left( {{D}^{2}}+4D+13 \right)y=\cos 3x$. First, we find out the PI differentiations.
We have $\left( {{D}^{2}}+4D+13 \right)y=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+4\dfrac{dy}{dx}+13y$.
Differentiating both side of $y\left( x \right)=A\cos 3x+B\sin 3x$, we get
$\begin{aligned} & y\left( x \right)=A\cos 3x+B\sin 3x \\\ & \Rightarrow \dfrac{dy}{dx}=-3A\sin 3x+3B\cos 3x \\\ \end{aligned}$
We differentiate again to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
$\begin{aligned} & \dfrac{dy}{dx}=-3A\sin 3x+3B\cos 3x \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-9A\cos 3x-9B\sin 3x \\\ \end{aligned}$
We put the values in the equation to get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+4\dfrac{dy}{dx}+13y \\\ & =\left( -9A\cos 3x-9B\sin 3x \right)+4\left( -3A\sin 3x+3B\cos 3x \right)+13\left( A\cos 3x+B\sin 3x \right) \\\ & =\cos 3x\left( 4A+12B \right)+\sin 3x\left( 4B-12A \right) \\\ \end{aligned}$
We have to satisfy the value of $\left( {{D}^{2}}+4D+13 \right)y=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+4\dfrac{dy}{dx}+13y$ with $\cos 3x$.
We equate them to get two equations of two unknowns A and B.
$\cos 3x\left( 4A+12B \right)+\sin 3x\left( 4B-12A \right)=\cos 3x$.
The equations are $\left( 4A+12B \right)=1,\left( 4B-12A \right)=0$.
We multiply the first equation with 3 and add it to the second equation.
$\begin{aligned} & 3\left( 4A+12B \right)=3,\left( 4B-12A \right)=0 \\\ & \Rightarrow 12A+36B=3,4B-12A=0 \\\ \end{aligned}$
Adding them we get
$\begin{aligned} & \left( 12A+36B \right)+\left( 4B-12A \right)=3+0 \\\ & \Rightarrow 40B=3 \\\ & \Rightarrow B=\dfrac{3}{40} \\\ \end{aligned}$
Putting value of B in one equation we get
$\begin{aligned} & \left( 4A+12B \right)=1 \\\ & \Rightarrow 4A+12\left( \dfrac{3}{40} \right)=1 \\\ & \Rightarrow 4A=1-\dfrac{9}{10}=\dfrac{1}{10} \\\ & \Rightarrow A=\dfrac{1}{40} \\\ \end{aligned}$
We got values of both A and B.
Putting the values in $y\left( x \right)=A\cos 3x+B\sin 3x$ we get $y\left( x \right)=\dfrac{\cos 3x}{40}+\dfrac{3\sin 3x}{40}$.
Simplifying we get $40y=\cos 3x+3\sin 3x$. This is the solution of $\left( {{D}^{2}}+4D+13 \right)y=\cos 3x$.

Note: We also can solve it by breaking it in two parts of CF and PI. Here CF defines the complementary function which is the solution of $\left( {{D}^{2}}+4D+13 \right)y=0$. Then we place the value of 3 in the particular integral. The final solution becomes $y\left( x \right)=CF+PI$.