solve the 1st order unforced differential equation

Jun 2016
16
0
london
solve the 1st order homogeneous differential equation

$2t{w}'-3w=0$

WORKED ANSWER

For a 1st order homogeneous ODE, have solution formula:

$A\cdot e^{-\int p}$

So we are doing the following integration

$e^{-\int \frac{3}{2t}}\cdot A$

$\Rightarrow \int 3\cdot \frac{1}{t}\cdot \frac{1}{2}dt$

$\Rightarrow \int \frac{3}{2}\cdot \frac{1}{t}dt$

$\Rightarrow \frac{3}2\cdot \ln t$

Now we've done the integration, sub-in to the original solution formula. Using laws of logarithms:

$A\cdot e^{\ln t^{-\frac{3}{2}}}$

$\Rightarrow A\cdot t\cdot -\frac{3}{2}$


PROBLEM

The actual solution given is:

$c\cdot t^{\frac{3}{2}}$


So you can see the solution and worked answer different by -1.


Any help please guys?
 
Last edited:

romsek

MHF Helper
Nov 2013
6,746
3,037
California
solve the 1st order homogeneous differential equation

$2t{w}'-3w=0$

WORKED ANSWER

For a 1st order homogeneous ODE, have solution formula:

$A\cdot e^{-\int p}$
no, it's $A \cdot e^{\int p}$
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
solve the 1st order homogeneous differential equation

$2t{w}'-3w=0$
$\displaystyle \begin{align*} 2\,t\,w' - 3\,w &= 0 \\ w' - \frac{3}{2\,t}\,w &= 0 \end{align*}$

This is first order linear, so the integrating factor is $\displaystyle \begin{align*} \mathrm{e}^{\int{-\frac{3}{2\,t}\,\mathrm{d}t}} = \mathrm{e}^{ -\frac{3}{2}\ln{(t)} } = \mathrm{e}^{ \ln{ \left( t^{-\frac{3}{2}} \right) } } = t^{-\frac{3}{2}} \end{align*}$, so multiplying both sides by the integrating factor gives

$\displaystyle \begin{align*} w' - \frac{3}{2\,t}\,w &= 0 \\ t^{-\frac{3}{2}}\,w' - \frac{3}{2}\,t^{-\frac{5}{2}}\,w &= 0 \\ \frac{\mathrm{d}}{\mathrm{d}t} \,\left( t^{-\frac{3}{2}}\,w \right) &= 0 \\ t^{-\frac{3}{2}}\,w &= \int{ 0\,\mathrm{d}t} \\ t^{-\frac{3}{2}} \, w &= C \\ w &= C\,t^{\frac{3}{2}} \end{align*}$
 
Jun 2016
16
0
london
no, it's $A \cdot e^{\int p}$
Incorrect.


The solution formula $A\cdot e^{-{\int p}}$ is derived as follows:

$2tw'-3w=0$

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

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

$\Rightarrow\ln w=-\frac{3}{2t}+c$

$\Rightarrow w=e^{-\int\frac{3}{2t}+C}$

$\Rightarrow A\cdot e^{-\int\frac{3}{2t}}$


The mistake I made was in the integration namely

$e^{-\int \frac{3}{2t}} \Rightarrow t^{-\frac{3}{2}}$


Thanks to "Prove It" for the help who solved the problem a different way, namely using integrating factors) is given by "Prove It".

Problem solved thank you