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I am working on a reduction of order problem and the last step is to solve what should be a simple first-order linear differential equation but frankly I'm not very good at them.

My problem started as: \begin{equation} 2tw'-3w = 0 \end{equation}

I recall that first order differential equations have the format \begin{equation} \frac{\mathop{}\!\mathrm{d}w}{\mathop{}\!\mathrm{d}t} + p(t)w = g(t) \end{equation} Thus, I set $p(t)$ to $-3$, and calculate the integrating factor $u(t)$: \begin{equation} e^{\int-3\mathop{}\!\mathrm{d}t} \implies e^{-3t} \end{equation}

Multiplying the whole equation by the integrating factor, I get \begin{equation} e^{-3t}2tw'-3e^{-3t}w = 0 \end{equation} Now, this is close to what I remember from normal linear DQs in that I can reduce this to the product rule (I see a $e^{-3t}$ and a $-3e^{-3t}$), but there's just one problem... what do I do about the $2t$ mashed in the middle?

For reference, the answer turns out to be $w(t) = ct^{3/2}$; I do not know how to get from where I am to that.

Isochronon
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Joe Balin
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2 Answers2

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You have to organize the differential equation into the form you are using (the coefficient of $w'$ shoulde be $1$),

$$ 2t w' -3w=0 \Rightarrow w' - \frac{3}{2t}w=0 $$

from this expression you can use $\frac{dw}{dt}+p(t)w=g(t) $, where $p(t) = -3/(2t)$

$$u(t) = e^{-\frac{3}{2} \ln t} = t^{-3/2}$$

multiplying this to the equation gives, $$ t^{-3/2}w'- \frac{3}{2}t^{-5/2} w = 0 $$ $$\Rightarrow (t^{-3/2}w)' =0 $$ $$ \Rightarrow t^{-3/2}w = c \Rightarrow w = c t^{3/2} $$

This is my first answer on math stackexcahnge for earning reputations!

mastrok
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This equation can be solved easier:

$$2t\frac{dw}{dt}=3w$$ $$\frac{dw}{w}=\frac{3}{2t}dt$$

After integration you get

$$\ln(w)=\ln(c)+\frac{3}{2}\ln(t)$$

Or

$$w=ct^{3/2}$$

MrYouMath
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