Answer to Question #107634 in Differential Equations for Nikesh gautam pandit ji

Question

Solve, using the method of variation of parameters
d^2y/dx^2 - y = 2/1+e^x
b) Solve the following initial value problem
d^2y/dx^2 + dy/DC -2y = -6sin2x -18cos2x
y(0)=2,y'(0)=2

Answer 1

a) The given equation can be written as

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

The auxiliary equation is

$m^2 - 1 = 0\\ m = \pm 1$

The complimentary function is

$C.F = c_{1}e^x + c_{2}e^{-x}= c_{1}f_{1} + c_{2}f_{2},$ where $f_{1} = e^x, f_{2} = e^{-x}$

The particular integral is = $Pf_{1}+Qf_{2}$, where

$P = -\displaystyle\int \dfrac{f_{2}g(x)}{W(f_{1},f_{2})}dx, Q = \displaystyle\int \dfrac{f_{1}g(x)}{W(f_{1},f_{2})}dx$

where, $g(x) = \dfrac{2}{1+e^{x}}\\$

$W(f_{1},f_{2}) = f_{1}f'_{2} - f'_{1}f_{2} = e^{x}(-e^{-x}) - e^{x}e^{-x} = -2$

$P = -\displaystyle\int \dfrac{2e^{-x}}{-2(1+e^x)}dx = \displaystyle\int \dfrac{e^{-x}}{1+e^{x}}dx\\=\displaystyle\int \dfrac{e^{-2x}}{1+e^{-x}}dx$

Put $u = 1+e^{-x}$ , then $du = -e^{-x}dx$

$dx = -\dfrac{dy}{y-1}$ .Therefore,

$P = \displaystyle\int \dfrac{(y-1)^2~(-dy)}{y(y-1)} = -\displaystyle\int \dfrac{y-1}{y}dy\\= \displaystyle\int (\dfrac{1}{y} - 1)dy = \ln y - y \\ P= \ln (1+e^{-x}) -1-e^{-x}$

$Pf_{1} = e^{x}(\ln(1+e^{-x}) - e^{-x}-1) \\= e^{x}(\ln(1+e^{-x})-1)-1$

$Q = \displaystyle\int \dfrac{2e^{x}}{-2(1+e^{x})}dx = -\displaystyle\int \dfrac{e^{x}}{1+e^{x}}dx\\ Q=-\ln(1+e^{x})\\ Qf_{2} = -e^{-x}\ln(1+e^{x})\\ \text{Hence,} \\ P.I = e^{x}(\ln(1+e^{-x})-1)- e^{-x}\ln(1+e^{x})-1.$

The complete solution is

$y = C.F.+P.I = e^{x}(c_{1} + \ln(1+e^{-x}) - 1) + e^{-x}(c_{2} - \ln(1+e^{x})) - 1$

b) The given equation can be written as, $(D^{2} + D - 2)y = -6\sin2x-18\cos2x, y(0) = 2,y'(0)=2.$

The auxiliary equation is,

$m^{2} + m -2 =0 \\(m+2)(m-1)=0\\ m = -2,1\\ C.F = c_{1}e^{-2x} + c_{2}e^{x}\\$

$P.I = \dfrac{1}{D^{2}+D-2}(-6\sin2x-18\cos2x)\\ =-6 \dfrac{1}{D^{2}+D-2}(\sin2x+3\cos2x)\\ =-6(\dfrac{1}{D^{2}+D-2}(\sin2x) + \dfrac{1}{D^{2}+D-2}(3\cos2x))\\ =-6(\dfrac{1}{-4+D-2}(\sin2x) + 3\dfrac{1}{-4+D-2}(\cos2x))~~(Replace~ D^{2}~ by -4)\\ =-6(\dfrac{1}{-6+D}(\sin2x) + 3\dfrac{1}{-6+D}(\cos2x))\\ =6(\dfrac{1}{6-D}(\sin2x) + 3\dfrac{1}{6-D}(\cos2x))\\ =6(\dfrac{6+D}{36-D^{2}}(\sin2x) + 3\dfrac{6+D}{36-D^{2}}(\cos2x))\\ =\dfrac{3}{20}(6+D)(\sin2x+3\cos2x)~~(Replace ~D^{2}~by -4)\\ =\dfrac{3}{20}(6\sin2x + 18\cos2x)+D(\sin2x+3\cos2x)\\ =\dfrac{3}{20}(6\sin2x + 18\cos2x +2\cos2x-6\sin2x)\\ P.I=3\cos2x$

The complete solution is

$y(x)= c_{1}e^{-2x}+c_{2}e^{x}+3\cos2x$

Given the initial conditions, $y(0)=2, y'(0)=2.$ We get,

$2=y(0)= c_{1}+c_{2}+3\\ c_{1}+c_{2} = -1\\ Now, y'(x) = -2c_{1}e^{-2x}+c_{2}e^{x}-6\sin2x\\ 2=y'(0) = -2c_{1}+c_{2}\\ -2c_{1}+c_{2} = 2\\$

Solving for $c_{1},c_{2}$ we get, $c_{1} = -1, c_{2} = 0.$

Thus, the complete solution becomes,

$y(x) = 3\cos2x-e^{-2x}$