Answer to Question #122595 in Differential Equations for jse

Question

1. Solve the given differential equation by undetermined coefficients.
y'' + 6y' + 9y = −xe^4x

y(x) =____

2. Solve the given differential equation by undetermined coefficients.
y''' − 3y'' + 3y' − y = e^x − x + 21

y(x)= _____

Answer 1

1.The given equation can be written as, $(D^{2}+6D+9)y = -xe^{4x}.$ The auxiliary equation is,

$m^2+6m+9=0.$ Solving we get, $m=-3(twice)$. The complementary function is,

$y_{c} = (c_{1}+c_{2}x)e^{-3x}$ . Let $y_{p} = (Ax+B)e^{4x}$ be the trial particular solution.

Then, $y_{p}$ satisfies the given differential equation. Therefore, $y''_{p}+6y'_{p}+9y_{p} = -xe^{4x}$ .

Now,

$\begin{aligned} y'_{p} &= e^{4x} \cdot A+ (Ax+B)\cdot e^{4x} \cdot 4\\ &= e^{4x}(4Ax+A+4B)\\ y''_{p}&= e^{4x} \cdot 4A + (4Ax+A+4B)\cdot e^{4x}\cdot 4\\ &= e^{4x}(16Ax+8A+16B) \end{aligned}$

Hence,

$\begin{aligned} e^{4x}\bigg(16Ax+8A+16B+6\cdot(4Ax+A+4B)+9 \cdot (Ax+B)\bigg) &= -xe^{4x}\\ 49Ax+14A+49B &= -x\\ \end{aligned}$

Comparing the coefficients and constant terms and solving we get,

$A = -\dfrac{1}{49}, B = \dfrac{2}{343}$ . Therefore, $y_{p} = \left(-\dfrac{x}{49}+\dfrac{2}{343}\right)e^{4x} = -\dfrac{e^{4x}}{49}\left(x-\dfrac{2}{7}\right)$ .

The general solution is,

$y = (c_{1}+c_{2}x)e^{-3x} - \dfrac{e^{4x}}{49}\left(x-\dfrac{2}{7}\right)$

2.The given equation can be written as, $(D^{3}-3D^{2}+3D-1)y=e^{x}-x+21$. The auxiliary equation is $m^{3}-3m^{2}+3m-1=0.$ Solving we get, $m = 1 (thrice).$ The complementary function is, $y_{c} = (c_{1}+c_{2}x+c_{3}x^{2})e^{x}.$ Since the complementary function is a linear combination of the term $e^{x}$, we choose the trial solution as $y_{p} = Ax^{3}e^{x} + Bx + C.$

Thus,

$\begin{aligned} y'_{p} &= A(x^{3}e^{x}+3x^{2}e^{x})+B = Ae^{x}(x^{3}+3x^{2})+B\\ y''_{p} &= A\left(e^{x}(3x^{2}+6x)+e^{x}(x^{3}+3x^{2})\right)\\ &= Ae^{x}\left(x^{3}+6x^{2}+6x\right)\\ y'''_{p} &= A\left(e^{x}(3x^{2}+12x+6)+e^{x}(x^{3}+6x^{2}+6x)\right)\\ &=Ae^{x}(x^{3}+9x^{2}+18x+6) \end{aligned}$

Using this in the given differential equation we get,

$\begin{aligned} e^{x}\bigg(Ax^{3}+9Ax^{2}+18Ax+6A\\ -3\cdot(Ax^{3}+6Ax^{2}+6Ax) \\ +3\cdot(Ax^{3}+3Ax^{2})-Ax^{3}\bigg) \\ +3B - Bx - C &= e^{x}-x+21\\ 6Ae^{x}-Bx+(3B-C)&=e^{x}-x+21 \end{aligned}$

Comparing the coefficients, constant terms and solving we get,

$A=\dfrac{1}{6}, B = 1, C = 18.$

Hence, $y_{p}= \dfrac{x^{3}e^{x}}{6}+x-18.$

The general solution is,

$y=(c_{1}+c_{2}x+c_{3}x^{2})e^{x}+\dfrac{x^{3}e^{x}}{6}+x-18.$