Question

Solve the given differential equation by undetermined coefficients.

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{cosx}\mathbf{-}\mathbf{sinx}$

Short answer

$y=C_1\cos x+C_2\sin x+\frac{1}{2}x\cos x+2x\sin x$

Step-by-step solution

Definition of complementary function

A complementary function is the general solution of a homogenous, linear differential equation.

Given data

From$m^2+1=0$ we have$m=\pm i$

So, the complementary function is$y_c=C_1\cos x+C_2\sin x$ .

Evaluation

From (5), of section 4.5 , with$\alpha =0,\beta =1$ and$n=1$

We have

$\left(D^2+1\right)\{\underset{\sin x}{\cos x}=0$

So,$\left(D^2+1\right)[4\cos x-\sin x]=0$

So, we apply the differential operator,$\left(D^2+1\right)$ to the both sides of the given differential equation, which gives

$\left(D^2+1\right)\left(D^2+1\right)y=\left(D^2+1\right)[4\cos x-\sin x]$

i.e,${\left(D^2+1\right)}^{2}y=0\cdots (*)$

Solve for particular solution

The auxiliary equation corresponding to (*) is

$$\begin{gathered} {\left(m^2+1\right)}^2=0 \\ \Rightarrow m_1=i=m_2,m_3=-i=m_4 \end{gathered}$$

Thus, the general solution must be .$y=\underset{y_c}{\underbrace{C_1\cos x+C_2\sin x}}+C_{3}x\cos x+C_{4}x\sin x$

Now, particular solution of the given differential equation should also satisfy (*) and the first two terms in y constitute $y_c$.

So, the remaining terms in y must be the basic form of $y_p$:

$y_p=Ax\cos x+Bx\sin x$

Substitution

Then$y_p^{\text{'}}=A\cos x+B\sin x-Ax\sin x+Bx\cos x$

$y_p^{*}=-(Ax\cos x+Bx\sin x)-2A\sin x+2B\cos x$

Now substituting $y_p$into the given differential equation gives

$-2A\sin x+2B\cos x=-\sin x+4\cos x$

On equating the coefficients of like terms, we get

$$\begin{gathered} -2A=-1 \\ 2B=4 \\ \therefore A=\frac{1}{2},B=2 \end{gathered}$$

So,
$y_p=\frac{1}{2}x\cos x+2x\sin x$

Thus, the required solution is .$y=C_1\cos x+C_2\sin x+\frac{1}{2}x\cos x+2x\sin x$