Question #299721

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

1
Expert's answer
2022-02-22T08:08:47-0500

Solution:

For complementary solution:

D25D+6=0D23D2D+6=0(D3)(D2)=0D=3,D=2D^2-5D+6=0 \\\Rightarrow D^2-3D-2D+6=0 \\\Rightarrow (D-3)(D-2)=0 \\\Rightarrow D=3,D=2

So, yc=c1e2x+c2e3xy_c=c_1e^{2x}+c_2e^{3x}

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

The particular solution to d2y(x)dx25dy(x)dx+6y(x)=xcos(2x)\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: yp(x)=a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x)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 a1,a2,a3, and a4a_{1}, a_{2}, a_{3} ,\ and\ a_{4}  :

Compute dyp(x)dx:dyp(x)dx=ddx(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))\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)

=2a1sin(2x)+a2cos(2x)2a2xsin(2x)+2a3cos(2x)+2a4xcos(2x)+a4sin(2x)\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 d2yp(x)dx2:\frac{d^{2} y_{p}(x)}{d x^{2}} :

d2yp(x)dx2=d2dx2(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))=4a1cos(2x)4a2xcos(2x)4a2sin(2x)4a3sin(2x)+4a4cos(2x)4a4xsin(2x)\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 yp(x)y_{p}(x)  into the differential equation:

d2yp(x)dx25dyp(x)dx+6yp(x)=xcos(2x)4a1cos(2x)4a2xcos(2x)4a2sin(2x)4a3sin(2x)+4a4cos(2x)4a4xsin(2x)5(2a1sin(2x)+a2cos(2x)2a2xsin(2x)+2a3cos(2x)+2a4xcos(2x)+a4sin(2x))+6(a1cos(2x)+a2xcos(2x)+a3sin(2x)+a4xsin(2x))=xcos(2x)\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:

(2a15a210a3+4a4)cos(2x)+(2a210a4)xcos(2x)+(10a14a2+2a35a4)sin(2x)+(10a2+2a4)xsin(2x)=xcos(2x)\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(2x)\cos (2 x)  on both sides of the equation:

2a15a210a3+4a4=02 a_{1}-5 a_{2}-10 a_{3}+4 a_{4}=0

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

2a210a4=12 a_{2}-10 a_{4}=1

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

10a14a2+2a35a4=010 a_{1}-4 a_{2}+2 a_{3}-5 a_{4}=0

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

10a2+2a4=010 a_{2}+2 a_{4}=0

Solve the system:

a1=5169a2=152a3=731352a4=552\begin{aligned} &a_{1}=-\frac{5}{169} \\ &a_{2}=\frac{1}{52} \\ &a_{3}=-\frac{73}{1352} \\ &a_{4}=-\frac{5}{52} \end{aligned}

Substitutea1,a2,a3, and a4 into yp(x)=cos(2x)a1+xcos(2x)a2+sin(2x)a3+xsin(2x)a4a_{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}  :

yp(x)=5169cos(2x)+152xcos(2x)73sin(2x)1352552xsin(2x)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:

y(x)=yc(x)+yp(x)=5169cos(2x)+152xcos(2x)73sin(2x)1352552xsin(2x)+c1e2x+c2e3x\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}


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