Solve the differential equation $t^2y''+3ty'+y=\frac{1}{t}$

Question

Solve the equation

$$t^2y''+3ty'+y=\frac{1}{t}$$ with $t \gt 0$

My try:

The given differential equation is actually of second order:

Let us use the substitution:

$ty=p$

$$ty'+y=\frac{dp}{dt}-(1)$$

$$ty''+2y'=\frac{d^2p}{dt^2}$$

$$t^2y''+2ty'=t\frac{d^2p}{dt^2}-(2)$$

Adding (1) and (2) we get:

$$t^2y''+3ty'+y=t\frac{d^2p}{dt^2}+\frac{dp}{dt}$$

Hence the equation is now:

$$t\frac{d^2p}{dt^2}+\frac{dp}{dt}=\frac{1}{t}$$

Let us use another substitution:

$\frac{dp}{dt}=q$

Then the equation is:

$$t\frac{dq}{dt}+q=\frac{1}{t}$$

$$\frac{dq}{dt}+\frac{q}{t}=\frac{1}{t^2}$$

Which is a Linear first order differential equation with integrating factor given by:

$$I(t)=e^{\int \frac{dt}{t}}=e^{\ln|t|}=|t|=t$$

The solution is:

$$qt=\int \frac{dt}{t}+C$$

$$qt=\ln t+C$$

$$q=\frac{\ln t+C}{t}$$

Now we get:

$$\frac{dp}{dt}=\frac{\ln t+C}{t}$$

Integrating we get:

$$p=\int \frac{ln t}{t}+\int \frac{C}{t}+D$$

$$p=\frac{1}{2}\left(\ln t\right)^2+C\ln(t)+D$$

Hence the final solution is:

$$ty=\frac{1}{2}\left(\ln t\right)^2+C\ln(t)+D$$

$$y=\frac{\frac{1}{2}\left(\ln t\right)^2+C\ln(t)+D}{t}$$

Where $C$ and $D$ are constants:

Is there any different approach?

Answer 1

The given ODE reduces to the following solution

\begin{align*} & t^{2}y^{\prime\prime} + 3ty^{\prime} + y = \frac{1}{t} \Leftrightarrow (t^{2}y^{\prime\prime} + 2ty^{\prime}) + (ty^{\prime} + y) = \frac{1}{t} \Leftrightarrow (t^{2}y^{\prime})^{\prime} + (ty)^{\prime} = \frac{1}{t} \Leftrightarrow\\ &\int (t^{2}y^{\prime})^{\prime}\mathrm{d}t + \int (ty)^{\prime}\mathrm{d}t = \int\frac{1}{t}\mathrm{d}t \Leftrightarrow t^{2}y^{\prime} + ty = \ln|t| + c \Leftrightarrow ty^{\prime} + y = \frac{\ln|t| + c}{t} \Leftrightarrow\\ & (ty)^{\prime} = \frac{\ln|t| + c}{t} \Leftrightarrow \int(ty)^{\prime}\mathrm{d}t = \int\left(\frac{\ln|t|}{t} + \frac{c}{t}\right)\mathrm{d}t \Leftrightarrow y(t) = \frac{\ln^{2}|t|}{2t} + \frac{c\ln|t|}{t} + \frac{k}{t} \end{align*}

Answer 2

The homogeneous equation is of Euler-Cauchy type, looking for basis solutions of the form $t^r$ leads to $$ 0=r(r-1)+3r+1=(r+1)^2. $$ The basis solutions for this double root are thus $t^{-1}$, $t^{-1}\ln t$.

As the right side has the same exponent $-1$, the method of undetermined coefficients provides indeed that the particular solution is of the form $y_p=Ct^{-1}(\ln t)^2$. Insertion into the equation allows then to compute $C$.