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# Solve the following differential equation; $(D^{2}+1)$Y=$0$ when $x=0;y=2$ and when $x=\large\frac{\pi}{2};$$y=-2 Can you answer this question? ## 1 Answer 0 votes Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X • Where X is a function of x • The solution is obtained in two parts. • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI obtained as y=\large\frac{1}{f(D)}$$\times h$
• The GS is $y=CF+PI$
• Let $m_1,m_2$ be the roots of the CE
• Case 1: $m_1,m_2$ are real numbers and distinct
• $CF=Ae^{m_1x}+Be^{m_2x}$
• Case 2: $m_1,m_2$ are complex (i.e)$m_1=\alpha+i\beta$ and $m_2=\alpha-i\beta$
• Then $CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x]$
• Case 3: $m_1,m_2$ are real and equal say $m_1$
• $CF=(A+Bx)e^{m_1x}$
• The PI when $X=e^{\alpha x},\alpha$ is a constant
• Case 1: $f(\alpha)\neq 0$
• $PI=\large\frac{1}{f(D)}e^{\alpha x}=\frac{e^{\alpha x}}{f(\alpha)}$
• Case 2: $f(\alpha)\neq 0$($\alpha=m_1,$ one of the roots of the CE)
• $\large\frac{1}{f(D)}e^{\alpha x}=\large\frac{1}{(D-\alpha)\theta(D)}e^{\alpha x}$ where $\theta(\alpha)\neq 0$
• $\Rightarrow \large\frac{1}{(D-\alpha)\theta(\alpha)}e^{\alpha x}=\large\frac{xe^{\alpha x}}{\theta(\alpha)}$
• Case 3: $f(\alpha)=0$ and $m_1=m_2=\alpha$ then
• $\large\frac{1}{f(D)}e^{\alpha x}=\large\frac{x^2}{2}e^{\alpha x}$
Step 1:
CE :$m^2+1=0\Rightarrow m=\pm i,\alpha =0,\beta =1$
CF : $A\cos x+B\sin x$
There is no P.I
Step 2:
GS :$y=A\cos x+B\sin x$
Step 3:
When $x=0,y=2$
$2=A$