Answer to Question #211149 in Differential Equations for Daniel

Question

Solve the following equations

1. d²y/dx² + 2dy/dx + 2y = e^x sin2x

2. d²y/dx² + 4dy/dx + 4y= x³e^2x

Answer 1

For differential equations of higher orders, we first need to make an auxiliary equation. Find roots of the auxiliary equation. These roots are used for finding Complementary Factor. Then find the Particular integral. Finally, add both CF and PI to give the solution of the equation.

(i) Given equation is: $\frac{d²y}{dx²} + 2\frac{dy}{dx} + 2y = e^x sin2x$

Auxiliary equation, $m^2+2m+2 = 0$

$m = -1+i, -1-i$

CF of the equation, $y = 2c_1e^{-x}cos(x)$

PI $=\frac{1}{D^2+2D+2}(e^x sin2x)$

Replace D with (D+1) then we get,

$=e^x \frac{1}{(D+1)^2+2(D+1)+2}sin2x = e^x \frac{1}{D^2+4D+5}sin2x$

Put -4 in place of $D^2$ in denominator,

$= e^x \frac{1}{-4+4D+5}sin2x = e^x \frac{1}{4D+1}sin2x = e^x \frac{4D-1}{(4D+1)(4D-1)}sin2x$

$=e^x \frac{4D-1}{(4D+1)(4D-1)}sin2x = e^x \frac{4D-1}{16D^2-1}sin2x$

$= e^x \frac{8cos2x-sin2x}{-16\times 4 - 1} = -\frac{1}{65}e^x ({8cos2x-sin2x})$

So solution will be, $y = 2c_1e^{-x}cos(x) -\frac{1}{65}e^x ({8cos2x-sin2x})$

(ii) Equation is, $\frac{d²y}{dx²} + 4\frac{dy}{dx} + 4y= x³e^2x$

Auxiliary equation, $m^2+4m+4 = 0$

$m = 2,2$

C.F., $y = e^{2x}(c_1+c_2x)$

P.I. $=\frac{1}{D^2+4D+4}x³e^{2x}$

Replace D with (D+2) then we get,

$= e^{2x}\frac{1}{(D+2)^2+4(D+2)+4}x^3 = e^{2x}\frac{1}{D^2+8D+16}x^3$

Expand denominator in term of D and its higher powers,

D means the differentiation of first order, and so on

$= \frac{e^{2x}}{16}(\frac{1}{1 + \frac{D^2+8D}{16}})x^3 = \frac{e^{2x}}{16}({1 + \frac{D^2+8D}{16}})^{-1}x^3$

$= \frac{e^{2x}}{16}({1 + \frac{D^2+8D}{16}})^{-1}x^3 = \frac{e^{2x}}{16}[1-(\frac{D^2+8D}{16})+(\frac{D^2+8D}{16})^2-......]x^3$

$= \frac{e^{2x}}{16} [1-(\frac{D^2+8D}{16})+(\frac{16D^3+64D^2}{256})-\frac{512D^3}{4096}]x^3$

neglecting higher-order as their derivative value will be zero.

$= \frac{e^{2x}}{16} [x^3 - \frac{6x+24x^2}{16}+\frac{96+384x}{256}-\frac{3072}{4096}]$

$= \frac{e^{2x}}{16}[x^3-\frac{3}{2}x^2-\frac{29}{128}x-\frac{3}{8}]$

So solution will be,

$y = e^{2x}(c_1+c_2x)+ \frac{e^{2x}}{16}[x^3-\frac{3}{2}x^2-\frac{29}{128}x-\frac{3}{8}]$