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How can I solve a 2nd order differential equation with non-constant coefficients like the following?

$$ty''-(t+1)y'+y=0$$

If I'm not wrong, I have only seen methods (apart from the reduction of order method) for finding a solution when the coefficients are constant. How can I do this?

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    $\begingroup$ Have you learned Laplace Transforms? $\endgroup$ – Amzoti Jan 21 '14 at 19:03
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Hint: The equation $ty''-(t+1) y'+y=0$ can be rearranged as:

$$ty''-(t+1) y'+y=0$$

$$0=ty''-(t+1)y'+y=ty''-ty'-y'+y=t(y''-y')-(y'-y)\\ \Leftrightarrow t(y''-y')=(y'-y)\\ \Leftrightarrow t\dfrac{d}{dt}(y'-y)=(y'-y).$$

Substituting $u=y'-y$, this becomes the 1st order ODE $$tu'=u\\ \Leftrightarrow\dfrac{u'}{u}=\dfrac{1}{t}\\ \Leftrightarrow\int\dfrac{u'}{u}dt=\int\dfrac{du}{u}=\int\dfrac{1}{t}dt\\ \Leftrightarrow\log{|u|}=\log{|t|}+C\\ \Leftrightarrow u(t)=c_1t$$

Once you solve for $u(t)$, the problem reduces to solving the linear 1st order ODE with constant coefficients, $y'-y=u(t)=c_1 t$.

$$y'-y=u(t)=c_1 t\\ \implies e^{-t}y'-e^{-t}y=u(t)=c_1 t e^{-t}\\ \implies \dfrac{d}{dt}\left(e^{-t}y\right)=c_1 t e^{-t}\\ \implies e^{-t}y = c_2 + c_1\int t e^{-t} dt\\ \implies e^{-t}y = c_2 + c_1 e^{-t}(t+1)\\ \implies y(t) = c_2 e^{t} + c_1 (t+1)$$

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  • $\begingroup$ Very nice solution! $\endgroup$ – Paul Safier Jan 21 '14 at 19:41
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If you use the Laplace transform technique, then you can reduce the order of the ode as

$$ (s-s^2)Y'(s) + (2-3s)Y(s) + c = 0, $$

where $Y(s)$ is the Laplace transform of $y(t)$. Now, solving the first order ode gives

$$ Y(s) = \frac{c_1\,s^2+c_2}{s^2(s-1)}. $$

Taking the inverse Laplace transform gives the solution of the original ode

$$ y(t) = A(t+1)+B e^{t}, $$

where $A, B$ are arbitrary constants.

Notes:

1) The Laplace transform is giving by

$$ Y(s) = \int_{0}^{\infty} y(t) e^{-st} dt. $$

2) To find the inverse Laplace transform, you can use partial fraction as

$$ Y(s) = -{\frac {c_{{2}}}{s}}+{\frac {c_{{1}}+c_{{2}}}{s-1}}-{\frac {c_{{2}}}{ {s}^{2}}}.$$

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