Question

Question: Solve the given boundary-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y=6\cos x-6\left(\cot 1\right)\sin x+x^2-1$

Step-by-step solution

Obtain the general solution of the homogeneous differential equation   

First, we need to solve the associated homogeneous equation $y\text{'}\text{'}+y=0$

Corresponding auxiliary equation is $m^2+1=0$

$(m^2-i^2)=0$

Solving,

$m_1=i$, $m_2=-i$, which are complex and conjugate.

So the solution of corresponding homogeneous equation is

$y_c=c_1\cos x+c_2\sin x$

Finding the particular solution

Given non homogeneous differential equation is $y\text{'}\text{'}+y=x^2+1$

Assume that $y_p=A{x}^2+Bx+C······\left(1\right)$is a solution of the non-homogeneous differential equation

Differentiate with respect to x

$$\begin{gathered} y_p\text{'}=2Ax+B······\left(2\right) \\ {y}_p\text{'}\text{'}=2A······\left(3\right) \end{gathered}$$

Substituting (1) and (3) in given differential equation

$2A+A{x}^2+Bx+C=x^2+1$

Comparing the coefficients,

$$\begin{gathered} A=1 \\ B=0 \\ 2A+C=1\Rightarrow C=-1 \end{gathered}$$

Hence the particular solution is$y_p=x^2-1$

Finding the solution

Hence the solution of given differential equation is

y = y_h + y_p

$y=C_1\cos x+C_2\sin x)+x^2-1······\left(4\right)$

Applying the given conditions

At the point (0,5)

(4) becomes $5=C_1\cos 0+C_2\sin 0+(0{)}^2-1$

Or $C_1=6$

At the point (1,0)

role="math" localid="1664170054814" $$\begin{gathered} 0=C_1\cos 1+C_2\sin 1+(1{)}^2-1 \\ {C}_2=-6\cot 1 \end{gathered}$$

Hence the solution is $\mathit{y}\mathbf{=}\mathbf{6}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{-}\mathbf{6}\left(cot1\right)\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$