2
$\begingroup$

I'm trying to find a general solution to: $$ \frac{d^2y}{dx^2} - 3\frac{dy}{dx} = 1+x $$

Firstly I've found the complementary function, by solving the homogeneous version of the D.E above, and I've got that it's: $$ y(x) = A + Be^{3x} $$

Now for the particular solution, I've tried $y=Px+Q \implies y'=P$ and $y''=0$, so subbing back into the D.E, I get: $$ -3P = 1 + x \\ P = -\frac{1}{3}-\frac{x}{3} $$ So $$y=(-\frac{1}{3}-\frac{x}{3})x = -\frac{x}{3}-\frac{x^2}{3}$$

Therefore my general solution is $$ y(x) = A + Be^{3x} -\frac{x}{3}-\frac{x^2}{3} $$

But this is not right, I'm not sure where I've gone wrong.

$\endgroup$
7
  • 3
    $\begingroup$ If you look for a particular solution $y(x)=Px+Q$, your $P$ cannot depend on $x$... Try rather $y(x)=ax^2+bx$ for the particular solution. $\endgroup$ – Did May 3 '14 at 11:01
  • $\begingroup$ @Did Is this because there is a constant in the complementary function and on the RHS of the non-homogeneous differential equation? $\endgroup$ – Michael May 3 '14 at 11:02
  • 1
    $\begingroup$ @Michael Yes. Suppose you just had the inhomogeneous equation $y''-3y'=1$. If you try to find a constant solution $y_p=c$ to the inhomogeneous equation, you get $c''-3c'=1$ which reduces to $0=1$. $\endgroup$ – David H May 3 '14 at 11:08
  • $\begingroup$ @DavidH Alright thanks, it makes more sense now :) $\endgroup$ – Michael May 3 '14 at 11:10
  • $\begingroup$ It helps to split the equation into two parts and find the $Y_p$ for each one... $y''-3y'=1$ and $ y''-3y=x$ $\endgroup$ – usukidoll May 3 '14 at 11:21
2
$\begingroup$

Since $0$ is a root for the characteristic equation then the particular solution take the form $$ax^2+bx^2+c$$ so substitute this polynomial in the differential equation and find the coefficients $a,b$ and $c$.

Remark

If we have an ODE $$ay''+by'+cy=P(x)e^{\alpha x}$$ then to find a particular solution there are three cases:

  • If $\alpha $ isn't a root to the characteristic equation so a particular solution to the ODE take the form $$Q(x)e^{\alpha x}$$ where $Q$ is a polynomial with $\deg Q=\deg P$.
  • If $\alpha $ is a simple root to the characteristic equation so a particular solution to the ODE take the form $$Q(x)e^{\alpha x}$$ where $Q$ is a polynomial with $\deg Q=\deg P+1$.
  • If $\alpha $ is a root to the characteristic equation with multiplicity $2$ so a particular solution to the ODE take the form $$Q(x)e^{\alpha x}$$ where $Q$ is a polynomial with $\deg Q=\deg P+2$.
$\endgroup$
2
  • $\begingroup$ Thanks, I will try that equation. $\endgroup$ – Michael May 3 '14 at 11:05
  • $\begingroup$ I edited my answer to give the general result. $\endgroup$ – user63181 May 3 '14 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.