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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 186
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Find the solution to the initial value problem.

y''(x)-y'(x)-2y(x)=cosx-sin2x;     y(0)=-720,    y'(0)=15

The initial solution to the differential equation is: y=-310cosx+320sin(2x)-110sinx-120cos(2x)

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

y''(x)-y'(x)-2y(x)=cosx-sin2x                     ......(1)

Write the homogeneous differential equation of the equation (1),

y''(x)-y'(x)-2y(x)=0

The auxiliary equation for the above equation,

m2-m-2=0

Step 2: Find the complementary solution of the given equation.

Solve the above equation,

m2-m-2=0m2-2m+m-2=0m(m-2)+1(m-2)=0(m-2)(m+1)=0

The root of an auxiliary equation is,

m1=2,   m2=-1

The complementary solution of the given equation is,

yc=c1e2x+c2e-x

Step 3:Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

yp(x)=Acosx+Bsin(2x)+Csinx+Dcos(2x)                         . .....(2)

Now find the first and second derivatives of the above equation,

yp'(x)=-Asinx+2Bcos(2x)+Ccosx-2Dsin(2x)yp''(x)=-Acosx-4Bsin(2x)-Csinx-4Dcos(2x)

Substitute the value of y(x),  y'(x) and yp''(t) the equation (1),

⇒y''(x)-y'(x)-2y(x)=cosx-sin2x⇒-Acosx-4Bsin(2x)-Csinx-4Dcos(2x)-[-Asinx+2Bcos(2x)+Ccosx-2Dsin(2x)]      -2[Acosx+Bsin(2x)+Csinx+Dcos(2x)]=cosx-sin2x⇒(-3A-C)cosx+(-3C+A)sinx+(-6D-2B)cos(2x)+(-6B+2D)sin(2x)=cosx-sin2x

Comparing all coefficients of the above equation,

-3A-C=1                             . .....(3)-6B+2D=-1                      . .....(4)-3C+A=0                          . .....(5)-6D-2B=0                        . .....(6)

Solve the equation (3) and (5),

role="math" localid="1655096234005" 3(-3A-C)=1×3-3C-9A=3 -3C+A=0               A=-310

Substitute the value of A in the equation (3),

role="math" localid="1655096019737" -3A-C=1-3-310-C=1c=910-1c=-110

Solve the equation (4) and (6),

role="math" localid="1655096203374" 3(-6B+2D)=−1×36D-18B=-3-6D-2B=0B=320

Substitute the value of B in the equation (4),

role="math" localid="1655096279255" -6B+2D=-1-6320+2D=-12D=-1+910D=-120

Substitute the value of A, B, C, and D in the equation (2),

yp(x)=Acosx+Bsin(2x)+Csinx+Dcos(2x)yp(x)=-310cosx+320sin(2x)-110sinx-120cos(2x)

Step 4: Find the general solution and use the given initial condition. 

Therefore, the general solution is,

y=yc(x)+yp(x)y=c1e2x+c2e-x-310cosx+320sin(2x)-110sinx-120cos(2x)             . ....(7)

Given the initial condition,

y(0)=-720,     y'(0)=15

Substitute the value of y=-720 and x = 0 in the equation (7),

-720=c1e2(0)+c2e-0-310cos(0)+320sin(0)-110sin(0)-120cos(0)-72=c1+c2-310-120c1+c2=-720+310+120c1+c2=0                                         . .....(8)

Now find the derivative of the equation (7),

y'=2c1e2x-c2e-x+310sinx+310cos(2x)-110cosx+110sin(2x)

Substitute the value of y'=15and x = 0 in the above equation,

15=2c1e2(0)-c2e-0+310sin(0)+310cos(0)-110cos(0)+110sin(0)15=2c1-c2+310-1102c1-c2=15-310+1102c1-c2=0                                                     . .....(9)

Solve the equation (8) and (9),

c1+c2=02c1-c2=0c1=0

Substitute the value of C1=0 in the equation (8),

c1+c2=0c2=0

Substitute the value of c1=0 and c2=0 in the equation (7),

role="math" localid="1655097899726" y=(0)e2x+(0)e-x-310cosx+320sin(2x)-110sinx-120cos(2x)y=-310cosx+320sin(2x)-110sinx-120cos(2x)

Thus, the initial solution to the differential equation is:

y=-310cosx+320sin(2x)-110sinx-120cos(2x)

Most popular questions for Math Textbooks

Find a particular solution to the given higher-order equation.

y'''-2y''-y'+2y=2t2+4t-9

Decide whether the method of undetermined coefficients together with superposition can be applied to find a particular solution of the given equation. Do not solve the equation.

2y''+3y'-4y=2t+sin2t+3

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) y''-y'-12y=2t6e-3t

The auxiliary equations for the following differential equations have repeated complex roots. Adapt the "repeated root" procedure of Section 4.2 to find their general solutions:

(a)y''''+2y''+y=0

(b)y''''+4y'''+12y''+16y'+16y=0

Find a general solution to the differential equation.

y''-y=-11t+1
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