Answer to Question #299721 in Differential Equations for Jitu

Question

(d^2-5d+6)y =x cos(2x)

Answer 1

Solution:

For complementary solution:

$D^2-5D+6=0 \\\Rightarrow D^2-3D-2D+6=0 \\\Rightarrow (D-3)(D-2)=0 \\\Rightarrow D=3,D=2$

So, $y_c=c_1e^{2x}+c_2e^{3x}$

Next, we determine particular solution by method of undetermined coefficients.

The particular solution to $\frac{d^{2} y(x)}{d x^{2}}-5 \frac{d y(x)}{d x}+6 y(x)=x \cos (2 x)$  is of the form: $y_{p}(x)=a_{1} \cos (2 x)+a_{2} x \cos (2 x)+a_{3} \sin (2 x)+a_{4} x \sin (2 x)$

Solve for the unknown constants $a_{1}, a_{2}, a_{3} ,\ and\ a_{4}$  :

Compute $\frac{d y_{p}(x)}{d x} : \frac{d y_{p}(x)}{d x}=\frac{d}{d x}\left(a_{1} \cos (2 x)+a_{2} x \cos (2 x)+a_{3} \sin (2 x)+a_{4} x \sin (2 x)\right)$

$\begin{aligned} =&-2 a_{1} \sin (2 x)+a_{2} \cos (2 x)-\\ & 2 a_{2} x \sin (2 x)+2 a_{3} \cos (2 x)+2 a_{4} x \cos (2 x)+a_{4} \sin (2 x) \end{aligned}$

Compute $\frac{d^{2} y_{p}(x)}{d x^{2}} :$

$\begin{aligned} \frac{d^{2} y_{p}(x)}{d x^{2}}=& \frac{d^{2}}{d x^{2}}\left(a_{1} \cos (2 x)+a_{2} x \cos (2 x)+a_{3} \sin (2 x)+a_{4} x \sin (2 x)\right) \\ =&-4 a_{1} \cos (2 x)-4 a_{2} x \cos (2 x)-\\ & 4 a_{2} \sin (2 x)-4 a_{3} \sin (2 x)+4 a_{4} \cos (2 x)-4 a_{4} x \sin (2 x) \end{aligned}$

Substitute the particular solution $y_{p}(x)$  into the differential equation:

$\begin{aligned} &\frac{d^{2} y_{p}(x)}{d x^{2}}-5 \frac{d y_{p}(x)}{d x}+6 y_{p}(x)=x \cos (2 x) \\ &-4 a_{1} \cos (2 x)-4 a_{2} x \cos (2 x)-4 a_{2} \sin (2 x)-4 a_{3} \sin (2 x)+ \\ &4 a_{4} \cos (2 x)-4 a_{4} x \sin (2 x)-5\left(-2 a_{1} \sin (2 x)+a_{2} \cos (2 x)-\right. \\ &\left.2 a_{2} x \sin (2 x)+2 a_{3} \cos (2 x)+2 a_{4} x \cos (2 x)+a_{4} \sin (2 x)\right)+ \\ &6\left(a_{1} \cos (2 x)+a_{2} x \cos (2 x)+a_{3} \sin (2 x)+a_{4} x \sin (2 x)\right)=x \cos (2 x) \end{aligned}$

Simplify:

$\begin{aligned} &\left(2 a_{1}-5 a_{2}-10 a_{3}+4 a_{4}\right) \cos (2 x)+\left(2 a_{2}-10 a_{4}\right) x \cos (2 x)+ \\ &\quad\left(10 a_{1}-4 a_{2}+2 a_{3}-5 a_{4}\right) \sin (2 x)+\left(10 a_{2}+2 a_{4}\right) x \sin (2 x)=x \cos (2 x) \end{aligned}$

Equate the coefficients of $\cos (2 x)$  on both sides of the equation:

$2 a_{1}-5 a_{2}-10 a_{3}+4 a_{4}=0$

Equate the coefficients of $x \cos (2 x)$  on both sides of the equation:

$2 a_{2}-10 a_{4}=1$

Equate the coefficients of $\sin (2 x)$  on both sides of the equation:

$10 a_{1}-4 a_{2}+2 a_{3}-5 a_{4}=0$

Equate the coefficients of $x \sin (2 x)$  on both sides of the equation:

$10 a_{2}+2 a_{4}=0$

Solve the system:

$\begin{aligned} &a_{1}=-\frac{5}{169} \\ &a_{2}=\frac{1}{52} \\ &a_{3}=-\frac{73}{1352} \\ &a_{4}=-\frac{5}{52} \end{aligned}$

Substitute$a_{1}, a_{2}, a_{3} ,\ and\ a_{4}\ into\ y_{p}(x)=\cos (2 x) a_{1}+x \cos (2 x) a_{2}+\sin (2 x) a_{3}+ x \sin (2 x) a_{4}$  :

$y_{p}(x)=-\frac{5}{169} \cos (2 x)+\frac{1}{52} x \cos (2 x)-\frac{73 \sin (2 x)}{1352}-\frac{5}{52} x \sin (2 x)$

The general solution is:

Answer:

$\begin{aligned} &y(x)=y_{\mathrm{c}}(x)+y_{p}(x)=-\frac{5}{169} \cos (2 x)+ \\ &\frac{1}{52} x \cos (2 x)-\frac{73 \sin (2 x)}{1352}-\frac{5}{52} x \sin (2 x)+c_{1} e^{2 x}+c_{2} e^{3 x} \end{aligned}$