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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''+y'-12y=et+e2t-1;     y(0)=1,    y'(0)=3

The solution to the initial value problem is:


See the step by step solution

Step by Step Solution

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

The differential equation is,

y''+y'-12y=et+e2t-1                      . .....(1)

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


The auxiliary equation for the above equation,


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

The root of an auxiliary equation is,

m1=3,   m2=-4

The complementary solution of the given equation is,


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

Assume, the particular solution of equation (1),

yp(t)=Aet+Be2t+C               . ....(2)

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


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


Comparing all coefficients of the above equation,

role="math" localid="1655098651285" -10A=1   A=-110-6B=1      B=-16-12C=-1C=112

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


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

Therefore, the general solution is,

y=yc(t)+yp(t)y=c1e3t+c2e-4t-110et-16e2t+112     . ....(3)

Given the initial condition,

y(0)=1,    y'(0)=3

Substitute the value of y = 1 and t = 0 in the equation (3),

role="math" localid="1655099043521" y=c1e3t+c2e-4t-110et-16e2t+1121=c1e3(0)+c2e-4(0)-110e0-16e2(0)+1121=c1+c2-110-16+112c1+c2=7160                                            . .....(4)

Now find the derivative of the equation (3),

role="math" localid="1655099101150" y'=3c1e3t-4c2e-4t-110et-13e2t

Substitute the value of y’ = 3 and t = 0 in the above equation,

role="math" localid="1655099214255" y'=3c1e3t-4c2e-4t-110et-13e2t3=3c1e3(0)-4c2e-4(0)-110e0-13e2(0)3=3c1-4c2-110-133c1-4c2=3+13303c1-4c2=10330                                     . .....(5)

Solve the equation (4) and (5),

role="math" localid="1655099327421" 4(c1+c2)=7160×44c1+4c2=71153c1-4c2=10330                c1=3530                c1=76

Substitute the value of c1in the equation (4),

role="math" localid="1655099442348" c1+c2=716076+c2=7160c2=7160-76c2=160

Substitute the value of c1 and c2 in the equation (3),

role="math" localid="1655099619782" y=c1e3t+c2e-4t-110et-16e2t+112y=76e3t+160e-4t-110et-16e2t+112

Thus, the solution to the initial value problem is:

role="math" localid="1655099676357" y=76e3t+160e-4t-110et-16e2t+112

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