# Solve the following differential equation;$(D^{2}-1)$$y=\cos$$2x-2\sin$$2x ## 1 Answer Need homework help? Click here. Toolbox: • A general second-order homogeneous equation is of the form a\large\frac{d^2y}{dx^2}$$+b\large\frac{dy}{dx}$$+cy=X • Where X is a function of x • The solution is obtained in two parts. • The first part is the complementary function CF • This is obtained by solving the equation am^2+bm+c=0.The second part is called the particular integral or PI • The GS is y=CF+PI • Let m_1,m_2 be the roots of the CE • Case 1: m_1,m_2 are real numbers and distinct • CF=Ae^{m_1x}+Be^{m_2x} • Case 2: m_1,m_2 are complex (i.e)m_1=\alpha+i\beta and m_2=\alpha-i\beta • Then CF=e^{\alpha x}[A\cos\beta x+B\sin \beta x] • Case 3: m_1,m_2 are real and equal say m_1 • PI :When x is of the form \cos\alpha x or \sin\alpha x • Case 1 : When f is a function of D^2 • PI =\large\frac{1}{f(D)}$$\cos\alpha x(or \sin\alpha x)=\large\frac{1}{\phi(D^2)}$$\cos\alpha x=\large\frac{1}{\phi(-\alpha ^2)}$$\cos\alpha x$
• Case 2:When $f=\phi(D,D^2)$
• PI is obtained by replacing $D^2$ by $-\alpha^2$
• $PI=\large\frac{1}{-\alpha ^2+bD+1}$$\cos\alpha x(or \sin\alpha x) • \;\;\;=\large\frac{1}{bD+P}$$\cos\alpha x(or \sin\alpha x)$
• $\;\;\;=\large\frac{bD-P}{(bD+P)(bD-P)}$$\cos\alpha x(or \sin\alpha x) • \;\;\;=\large\frac{bD-P}{(bD^2-P^2)}$$\cos\alpha x(or \sin\alpha x)$
• $\;\;\;=\large\frac{bD-P}{(b(-\alpha^2)-P^2)}$$\cos\alpha x(or \sin\alpha x) • The denominator is a constant and the numerator represents a differentiation • Case 3: If \phi(-\alpha^2)=0 then • PI=\large\frac{1}{\phi(D^2)}$$\cos \alpha x=\large\frac{1}{D^2+\alpha^2}$$\cos\alpha x • \Rightarrow R.P of \large\frac{1}{(D+i\alpha)(D-i\alpha)}$$e^{i\alpha x}$
• $\Rightarrow$ R.P of $\large\frac{1}{\theta(i\alpha)}$$e^{i\alpha x} • \Rightarrow \large\frac{-x}{2\alpha}$$(-\sin \alpha x)=\large\frac{x\sin\alpha x}{2\alpha}$
• or $PI=\large\frac{1}{\phi(D^2)}$$\sin\alpha x=I.P of \large\frac{1}{(D+i\alpha)(D-i\alpha)}$$e^{i\alpha x}$
• PI=$\large\frac{-x}{2\alpha}$$\cos\alpha x Step 1: CE : m^2-1=0 (m+1)(m-1)=0 m_1=-1,m_2=1 CF=Ae^{-x}+Be^x Step 2: PI =\large\frac{1}{D^2-1}$$\cos 2x-\large\frac{1}{D^2-1}$$2\sin 2x Step 3: PI_1=\large\frac{1}{D^2-1}$$\cos 2x=\large\frac{1}{-4-1}$$\cos 2x=\large\frac{-1}{5}$$\cos 2x$
Step 4: