Question #107634
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
1
Expert's answer
2020-04-06T16:29:32-0400

a) The given equation can be written as

(D21)y=21+ex(D^2 - 1)y = \frac{2}{1+e^x}

The auxiliary equation is

m21=0m=±1m^2 - 1 = 0\\ m = \pm 1

The complimentary function is

C.F=c1ex+c2ex=c1f1+c2f2,C.F = c_{1}e^x + c_{2}e^{-x}= c_{1}f_{1} + c_{2}f_{2}, where f1=ex,f2=exf_{1} = e^x, f_{2} = e^{-x}


The particular integral is = Pf1+Qf2Pf_{1}+Qf_{2}, where

P=f2g(x)W(f1,f2)dx,Q=f1g(x)W(f1,f2)dxP = -\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)=21+exg(x) = \dfrac{2}{1+e^{x}}\\


W(f1,f2)=f1f2f1f2=ex(ex)exex=2W(f_{1},f_{2}) = f_{1}f'_{2} - f'_{1}f_{2} = e^{x}(-e^{-x}) - e^{x}e^{-x} = -2

P=2ex2(1+ex)dx=ex1+exdx=e2x1+exdxP = -\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+exu = 1+e^{-x} , then du=exdxdu = -e^{-x}dx

dx=dyy1dx = -\dfrac{dy}{y-1} .Therefore,


P=(y1)2 (dy)y(y1)=y1ydy=(1y1)dy=lnyyP=ln(1+ex)1exP = \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}

Pf1=ex(ln(1+ex)ex1)=ex(ln(1+ex)1)1Pf_{1} = e^{x}(\ln(1+e^{-x}) - e^{-x}-1) \\= e^{x}(\ln(1+e^{-x})-1)-1


Q=2ex2(1+ex)dx=ex1+exdxQ=ln(1+ex)Qf2=exln(1+ex)Hence,P.I=ex(ln(1+ex)1)exln(1+ex)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=ex(c1+ln(1+ex)1)+ex(c2ln(1+ex))1y = 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, (D2+D2)y=6sin2x18cos2x,y(0)=2,y(0)=2.(D^{2} + D - 2)y = -6\sin2x-18\cos2x, y(0) = 2,y'(0)=2.

The auxiliary equation is,

m2+m2=0(m+2)(m1)=0m=2,1C.F=c1e2x+c2exm^{2} + m -2 =0 \\(m+2)(m-1)=0\\ m = -2,1\\ C.F = c_{1}e^{-2x} + c_{2}e^{x}\\

P.I=1D2+D2(6sin2x18cos2x)=61D2+D2(sin2x+3cos2x)=6(1D2+D2(sin2x)+1D2+D2(3cos2x))=6(14+D2(sin2x)+314+D2(cos2x))  (Replace D2 by4)=6(16+D(sin2x)+316+D(cos2x))=6(16D(sin2x)+316D(cos2x))=6(6+D36D2(sin2x)+36+D36D2(cos2x))=320(6+D)(sin2x+3cos2x)  (Replace D2 by4)=320(6sin2x+18cos2x)+D(sin2x+3cos2x)=320(6sin2x+18cos2x+2cos2x6sin2x)P.I=3cos2xP.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)=c1e2x+c2ex+3cos2xy(x)= c_{1}e^{-2x}+c_{2}e^{x}+3\cos2x

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

2=y(0)=c1+c2+3c1+c2=1Now,y(x)=2c1e2x+c2ex6sin2x2=y(0)=2c1+c22c1+c2=22=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 c1,c2c_{1},c_{2} we get, c1=1,c2=0.c_{1} = -1, c_{2} = 0.

Thus, the complete solution becomes,

y(x)=3cos2xe2xy(x) = 3\cos2x-e^{-2x}


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