0
$\begingroup$

Find the general solution to the following ODE:

$$y''-\frac{3}{x}y'+\frac{3}{x^2}y=x^2e^x$$

My working so far: Clearly this is a second order, non-homogeneous equation. The equation follows the form: $$a(x)y''+b(x)y'+c(x)y=g(x)$$ Thus, the general solution for this can be written as: $$y=y_h+y_p$$ yh:

yh is given by the solution to the homogeneous ODE a(x)y''+b(x)y'+c(x)y=0. So,

$$y''-\frac{3}{x}y'+\frac{3}{x^2}y=0$$ $$r^2-\frac{3}{x}r+\frac{3}{x^2}=0$$

The problem is that I don't know how to find the values for r

I am also confused about how to find yp (the particular solution).

Any help would be much appreciated. Thanks in advance.

$\endgroup$
1
  • $\begingroup$ It's Cauchy Euler's equation $\endgroup$ Mar 26, 2020 at 1:54

3 Answers 3

1
$\begingroup$

$$y''-\frac{3}{x}y'+\frac{3}{x^2}y=x^2e^x$$ multiply by $x^2$ $$x^2y''-3xy'+3y=x^4e^x$$ this is Cauchy-Euler Equation see

$\endgroup$
3
  • $\begingroup$ Thanks for the help, but I'm still unsure where to go with this. If I try and find yh using the homogenous equation I still don't know how to find r. $\endgroup$ Mar 26, 2020 at 1:46
  • $\begingroup$ Oh, i've got it now, thanks for the help. $\endgroup$ Mar 26, 2020 at 1:57
  • $\begingroup$ you're welcome. $\endgroup$
    – E.H.E
    Mar 26, 2020 at 2:04
1
$\begingroup$

Your eqaution has not constant coefficients so that method won't work. Note that:

$$y''-\frac{3}{x}y'+\frac{3}{x^2}y=x^2e^x$$ $$ \frac{y''}{x^2} - \frac{2y'}{x^3} - \frac{y'}{x^3} + \frac{3y}{x^4} =e^x$$ $$\left ( \frac{y'}{x^2} \right)'- \left ( \frac{y}{x^3} \right)'=e^x$$ Integrate. $$\left ( \frac{y'}{x^2} \right)- \left ( \frac{y}{x^3} \right)=e^x+c_1$$ $$\left ( \frac{xy'-y}{x^2} \right)=xe^x+c_1x$$ $$\left ( \frac{y}{x} \right)'=xe^x+c_1x$$ Integrate $$y=xe^x(x-1)+cx^3+c_2x$$

$\endgroup$
1
$\begingroup$

$$x^2Y''-3xY'+3Y=x^2e^{x}=f(x)~~~(1)$$ First solve the homogeneous part $$x^2y''-3xy'+3y=0.~~~~(2)$$ Let $y=x^m$ $$m(m-1)-3m+3=0 \implies m^2-4m+3 \implies m=3,1,$$ So the solutions are $y_1=C_1 x, y_2=C_2 x^3$ Next we vary $C_1,C_2$ as $C_1(x), C_2(x)$. $$Y(x)=C_1(x) x+C_2(x) x^3~~~(3)$$ The wronskian of $w(x)=y_1,y'_2-y'_1 y_2=w=2x^3$ where $$C_1(x)=-\int \frac{x^2y_2(x) f(x)}{w(x)} dx+D_1=\frac{1}{2}\int x^4 e^x dx+D_1$$ $$=-\frac{1}{2}[(x^4-4x^3+12x^2-24x+24)]e^x+D_1$$ $$C_2(x)=\int x^2 \frac{y_1(x) f(x)}{w(x)}dx+D_2=\frac{1}{2} \int x^2e^x dx+ D_2=\frac{(x^2-2x+2)e^x}{2}+D_2$$ Putting them in (3). we get the total solution of (1).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .