 osteoblogda

2021-12-31

Using the method of undetermined coefficients, find the general solution of the following differential equation ${y}^{″}-3{y}^{\prime }+2y={x}^{2}$. habbocowji

We find the general solution of eq (1) of the method of undesermened coefficient.
Now the aurilary eq for the homogenous eq
$y3{y}^{\prime }+2y=0$ is
${m}^{2}-3m+2=0$
$\left(m-2\right)\left(m-1\right)=0$
$m=1,2$
So the function $y={c}_{1}e+{c}_{2}{e}^{2x}$
Here we take ${y}_{p}=A{x}^{2}+Bx+c$,
${y}_{p}=2a+b$
${y}_{p}2a$
Putting the value of ${y}_{p},{y}_{p},{y}_{p}$ in eq (1) we get
$2a-3\left(2ax+b\right)+2\left(a{x}^{2}+bx+c\right)={x}^{2}$
$2a-6ax-3b+2a{x}^{2}+2bx+2={x}^{2}$
$2a{x}^{2}+\left(-6a+2b\right)x+\left(2a-3b+2c\right)={x}^{2}$
$2a=1⇒a=\frac{1}{2}$
$-6A+2b=0⇒-6×\frac{1}{2}+2b=0⇒-3+2b=0⇒b=\frac{3}{2}$
$2a-3b+2c=0⇒2\cdot \frac{1}{2}-3\cdot \frac{3}{2}+2=0$
$⇒1-\frac{9}{2}+2c=0$
$⇒2c\cdot \frac{9}{2}-1$
$2c=\frac{1}{2}$
$=1c=\frac{1}{4}$
${y}_{p}=\frac{1}{2}{x}^{2}+\frac{3}{2}x+\frac{1}{4}$
The general solution is
$y\left(x\right)={y}_{c}+{y}_{p}$
$={c}_{1}{e}^{x}+{c}_{2}{e}^{2x}+\frac{1}{2}{x}^{2}+\frac{3}{2}x+\frac{1}{4}$ Deufemiak7

Given differential equation is
$y3{y}^{\prime }+2y={x}^{2}+x+1$ (1)

$\left(D-1\right)\left(D-2\right)=0$
$\therefore D=1,2$
$\therefore CF={c}_{1}{e}^{x}+{c}_{2}{e}^{2x}$
PI is of the form
$y={A}_{0}+{A}_{1}x+{A}_{2}{x}^{2}$
${y}^{\prime }=0+{A}_{1}+2{A}_{2}x$
$y0+2{A}_{2}$
Substituting in (1)
$2{A}_{2}-3\left({A}_{1}+2{A}_{2}x\right)+2\left({A}_{0}+{A}_{1}x+{A}_{2}{x}^{2}\right)={x}^{2}+x+1$
$\therefore 2{A}_{2}{x}^{2}+\left(2{A}_{1}-6{A}_{2}\right)x+2{A}_{0}-3{A}_{1}+2{A}_{2}={x}^{2}+x+1$
Comparing the coefficients
$2{A}_{2}=1,2{A}_{1}-6{A}_{2}=1,2{A}_{0}-3{A}_{1}+2{A}_{2}=1$
Solving, ${A}_{2}=\frac{1}{2},{A}_{1}=2,{A}_{0}=3$
$\therefore PI=3+2x+\frac{1}{2}{x}^{2}$
$\therefore GS=CF+PI$
$y={c}_{1}{e}^{x}+{c}_{2}{e}^{2x}+\left(3+2x+\frac{1}{2}{x}^{2}\right)$ Vasquez

Solve ${y}^{″}-3{y}^{\prime }+2y={x}^{2}$
Homogen solution: $y=A{e}^{x}+B{e}^{2x}$
Particular solution:
${y}_{p}=x\left(A{x}^{2}+Bx+C\right)=A{x}^{3}+B{x}^{2}+Cx$
${y}^{\prime }p=3A{x}^{2}+2Bx+C$
${y}^{″}p=6Ax+2B$
Put his into the initial equartion to get A, B and C gives me:
$A=0,B=1/2,C=3/2$
$y=A{e}^{x}+B{e}^{2x}+{x}^{2}/2+3x/2$
$y=A{e}^{x}+B{e}^{2x}+{x}^{2}/2+3x/2+7/4$