Particular solution to y″−3y′+2y=2ex

Name: Particular solution to y″−3y′+2y=2ex
Uploaded: 2022-02-23T17:22:57+00:00
Description: Particular solution to y″−3y′+2y=2ex

Question

limacarp4 · Accepted Answer

y″−3y′+2y=2ex.Use operator D:Let D=ddx.So, (D2−3D+2)yp=2ex⇒(D−1)(D−2)yp=2ex⇒yp=1(D−1)(D−2)⋅2ex=2ex1(D+1−1)(D+1−2)⋅1=2ex1D(D−1)⋅1⇒yp=2ex1D(0−1)⋅1=−2ex1D⋅1=−2xex

ambarakaq8 · Accepted Answer

This is too long for a comment, so I posted it as an answer. First solve for the homogeneous equation $y^{″} - 3 y^{'} + 2 y = 0$ by setting the right hand side to be zero. The auxiliary equation is $m^{2} - 3 m + 2 = 0$ , which has roots $m = 2, 1$ . Therefore the solution for this homogeneous equation is $e^{x} and e^{2 x}$ . Now we want to find a particular solution $y^{″} - 3 y^{'} + 2 y = 2 e^{x}$ . Normally we set the particular solution to be $A e^{x}$ . However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with x until no duplication occurs. Therefore, the particular solution is given by $A x e^{x}$ .
Let me do another example: to solve $y^{″} - 2 y^{'} + y = 2 e^{x}$ .
First solve the homogenous equation $y^{″} - 2 y^{'} + y = 0$ . The auxiliary equation is $m^{2} - 2 m + 1 = 0$ which has double roots $m = 1$ . Therefore, the solution for this homogeneous equation is $e^{x} and x e^{x}$ . Now if we want to find a particular solution $y^{″} - 2 y^{'} + y = 2 e^{x}$ . Normally we set the particular solution to be $A e^{x}$ . However, it duplicates with the solution of the homogeneous solution, therefore, we multiple it with x and it becomes $A x e^{x}$ , but it still duplicates with $x e^{x}$ . Therefore, we mupltiply it by $x^{2}$ , and the particular solution is given by $A x^{2} e^{x}$ .

RizerMix · Accepted Answer

The simplest thing in your case is probably to use this (easily justified) trick: If p(x) is a one variable polynomial with complex coefficients and c a complex number, then the ODE p(ddx)y=ecx has a solution of the form ecxq(x) where q(x) is a polynomial whose degree is the number of roots of p(x) equal to c. In you case, p(x)=(x−1)(x−2),c=1, so the trick tells you to look for a solution of the form (ax+b)ex, and you can clearly assume b=0.

Particular solution to y''-3y'+2y=2e^{x}

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