Question

Solve the differential equation x^2y'' -3xy' +3y=2x^4e^x

Answer 1

The solution of the given differential equation is $y_{p}$ = 2$x^{2} e^{x}$ - 2x$e^{x}$

Given that;

x^2y'' -3xy' +3y=2x^4e^x

To find;

The solution of  the differential equation x^2y'' -3xy' +3y=2x^4e^x

Solution;

The given equation is an Euler-Cauchy differential equation. Find the characteristic polynomial given y=$x^{r}$ to solve the homogenous equation.

Let y = $x^{r}$, $y^{'}$ = r$x^{r}$ , $y^{'}$ = r(r - 1)$x^{r - 2}$

$x^{2}$$y^{"}$ - 3x$y^{'}$ + 3y = 0

$x^{2}$r(r - 1)$x^{r - 2}$ -3xr$x^{r - 1}$ + 3$x^{r}$ = 0

r(r - 1) -3r + 3 = 0 ⇔ $r^{2}$ - 4r + 3 = 0 = (r -1)(r - 3)

$y_{h}$ = A$x^{3}$ + Bx is the homogenous equation.

Next, we will find a particular solution using the method of undetermined coefficients.

$y_{p}$ = a$x^{2}$$e^{x}$ + bx$e^{x}$

$y^{'} _{p}$ = a$x^{2}$$e^{x}$ + 2ax$e^{x}$ + bx$e^{x}$ + b$e^{x}$

$y^{"} _{p}$ = a$x^{2}$$e^{x}$ + 4ax$e^{x}$ + 2a$e^{x}$ + bx$e^{x}$ + 2b$e^{x}$

multiply each by its coefficients in the ODE,

$x^{2}$$y^{" }$ - 3x$y^{'}$ + 3y = 2$x^{4}$$e^{x}$

3$y_{p}$ = 3a$x^{2}$$e^{x}$ + 3bx$e^{x}$

- 3x$y^{'} _{p}$ = - 3a$x^{3}$$e^{x}$ - 6a$x^{2}$$e^{x}$ - 3b$x^{2}$$e^{x}$ -3bx$e^{x}$

$x^{2}$$y^{"} _{p} }$ = a$x^{4}$$e^{x}$ +4a$x^{3} e^{x}$+ 2a$x^{2} e^{x}$ + b$x^{3} e^{x}$ + 2b$x^{2} e^{x}$4

match equal terms to determine a and b.

x$e^{x}$ : 3b - 3b = 0

b = b

$x^{2} e^{x}$ : 3a  - 6a - 3b + 2a + 2b = 0

a + b = 0

$x^{3} e^{x}$ : -3a + 4a + b = 0

b = - a

$x^{4} e^{x}$ : a =2

This over-determined linear system has a=2,b=(−2). The particular solution,

$y_{p}$ = 2$x^{2} e^{x}$ - 2x$e^{x}$

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