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Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1. Consider the differential equation d’y – 27y = 7e3x dr3 • State and give a general solution to the homogenous equation • Give a particular solution to the inhomogeneous equation • Give a general solution to the inhomogeneous equation
Transcribed Image Text:1. Consider the differential equation d’y – 27y = 7e3x dr3 • State and give a general solution to the homogenous equation • Give a particular solution to the inhomogeneous equation • Give a general solution to the inhomogeneous equation
i.e. y = Expected answer: A*e (2x) + B*e^(-2x/2) *cos (2x*sqrt(3)/2) + Cke^(-2x/2)*sin (2x*sqrt(3)/2) + 5x*e^(: 2xV Ae + Be# cos( 2) + ce# sin( 5xe2x + 12 Score: 0/1 X Unanswered Advice Give a general solution to y" – 8y = 5e2x Part a) First we find the complementary function, y, this is a general solution to the homogeneous equation y" - 8y = 0 We look for solutions of the form ye = Aekx wherek can be real, imaginary and complex. By substituting this into the differential equation we find that k satisfies k³ = 8. This has three solutions of the form ae where a = 2 and 0 = 0, , * 0 <0 < 2n, increasing order. In rectangular form these correspond to k = 2, k2 = -1+ iv3, kz = -1 - iv3. Part b) Thus the general solution to the equation is ye = Aekix + bek2* + ceksx rewriting this in terms of exponential and trigonometric functions we have y. = Ae2x + be* (cos(xv3) + i sin(xv3)) + ce* (cos (xV3) – i sin(xv3)) In order to restrict ourselves to real solutions we can take b and c to be complex conjugates of each other b = (B – iC)/2, c = b' = (B+ iC)/2 Substituting for these we have y. = Ae2x + Be-* cos(xv3) + Ce-* sin(xv3) Part c) Particular integral y, is a solution to the full equation y" - 8y, = 5e2x. Since e2x is a solution to the homogeneous equation we should try a trial solution of the form Yp = Dxe2* This expression has derivatives y, = 2Dxe2x + De²x y = 4Dxe2x + 4 De²x y = 8Dxe2x + 12D¢²* Substituting this into the equation we obtain 12 De2x = 5e2x, and thus D = Yp = S Part d) The general solution is y = ye + Yp, i.e. y = Ae2x + Be"* cos(xv3) + Ce-* sin(xv3) + x. Score: 0/22 X Try another question like this one
Transcribed Image Text:i.e. y = Expected answer: A*e (2x) + B*e^(-2x/2) *cos (2x*sqrt(3)/2) + Cke^(-2x/2)*sin (2x*sqrt(3)/2) + 5x*e^(: 2xV Ae + Be# cos( 2) + ce# sin( 5xe2x + 12 Score: 0/1 X Unanswered Advice Give a general solution to y" – 8y = 5e2x Part a) First we find the complementary function, y, this is a general solution to the homogeneous equation y" - 8y = 0 We look for solutions of the form ye = Aekx wherek can be real, imaginary and complex. By substituting this into the differential equation we find that k satisfies k³ = 8. This has three solutions of the form ae where a = 2 and 0 = 0, , * 0 <0 < 2n, increasing order. In rectangular form these correspond to k = 2, k2 = -1+ iv3, kz = -1 - iv3. Part b) Thus the general solution to the equation is ye = Aekix + bek2* + ceksx rewriting this in terms of exponential and trigonometric functions we have y. = Ae2x + be* (cos(xv3) + i sin(xv3)) + ce* (cos (xV3) – i sin(xv3)) In order to restrict ourselves to real solutions we can take b and c to be complex conjugates of each other b = (B – iC)/2, c = b' = (B+ iC)/2 Substituting for these we have y. = Ae2x + Be-* cos(xv3) + Ce-* sin(xv3) Part c) Particular integral y, is a solution to the full equation y" - 8y, = 5e2x. Since e2x is a solution to the homogeneous equation we should try a trial solution of the form Yp = Dxe2* This expression has derivatives y, = 2Dxe2x + De²x y = 4Dxe2x + 4 De²x y = 8Dxe2x + 12D¢²* Substituting this into the equation we obtain 12 De2x = 5e2x, and thus D = Yp = S Part d) The general solution is y = ye + Yp, i.e. y = Ae2x + Be"* cos(xv3) + Ce-* sin(xv3) + x. Score: 0/22 X Try another question like this one
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