# Math Help - Finding particular solution from differential equations

1. ## Finding particular solution from differential equations

I am stuck on how to solve the following, i don't weather my arithmetic is wrong or what i am doing. So could someone please do each step.
Find the particular solution to

a) dh/dt=2h(3-h) where y=1 and t=0
b) dy/dx= e^y +1/e^y where y=0 when x=0.

2. ## Re: Finding particular solution from differential equations

First of all try to find the general solution of the DE (it's a separable one), for example the first one:
$\frac{dh}{dt}=2h(3-h)$
Calculating general solution, first rewrite the DE:
$\frac{dh}{2h(3-h)}=dt$
Take the integral of both sides:
$\int \frac{dh}{2h(3-h)}=\int dt$
If you split the integrand in partial fractions (in LHS) you'll get:
$\frac{1}{6}\int \frac{dh}{h} - \frac{1}{6}\int \frac{dh}{h-3}=tt+C$
$=\frac{1}{6}\ln|h|-\frac{1}{6}\ln|h-3|=t+C$
$\Leftrightarrow \ln\left|\frac{h}{h-3}\right|=6t+6C$
$\Leftrightarrow e^{6t+6C}=\left|\frac{h}{h-3}\right|$

Now try to continue and try to get an expression $h(t)=...$ and afterwards substitute $h=1,t=0$ to find $C$.
(Note: $\ln$ is the natural logarithm)

3. ## Re: Finding particular solution from differential equations

Originally Posted by johnsy123
I am stuck on how to solve the following, i don't weather my arithmetic is wrong or what i am doing. So could someone please do each step.
Find the particular solution to

a) dh/dt=2h(3-h) where y=1 and t=0
b) dy/dx= e^y +1/e^y where y=0 when x=0.
Hiya Johnsy, is the second one $\displaystyle \frac{dy}{dx} = e^y + \frac{1}{e^y}$ or $\displaystyle \frac{dy}{dx} = \frac{e^y + 1}{e^y}$?

For the first...

\displaystyle \begin{align*} \frac{dh}{dt} &= 2h(3 - h) \\ \frac{dt}{dh} &= \frac{1}{2h(3-h)} \\ t &= \int{\frac{1}{2h(3 -h)}\,dh} \end{align*}

You will now need to apply partial fractions.

4. ## Re: Finding particular solution from differential equations

It is (e^y +1)/(e^y)

5. ## Re: Finding particular solution from differential equations

Originally Posted by johnsy123
It is (e^y +1)/(e^y)
OK, so

\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{e^y + 1}{e^y} \\ \frac{dx}{dy} &= \frac{e^y}{e^y + 1} \\ x &= \int{\frac{e^y}{e^y + 1}\,dy} \end{align*}

Now make the substitution $\displaystyle u = e^y + 1 \implies du = e^y\,dy$ and the integral becomes

$\displaystyle x = \int{\frac{1}{u}\,du}$.

I'm sure you can go from here

6. ## Be carefully using the results of Wolfram!...

Originally Posted by johnsy123
It is (e^y +1)/(e^y)
... so that the DE is...

$y^{'}= 1+e^{-y}\ ,\ y(0)=0$ (1)

In (1) the variables are separable and the solution is found with the standard approach...

$\int \frac{d y}{1+e^{-y}} = \int dx \implies y+ \ln (1+e^{-y}) = x + c$ (2)

... and the 'initial condition' means that is $c=0$. All right?... yes, of course, but it is interesting to spend a little of time about the details of the integration performed in (2). The integral is...

$\int \frac{d y}{1+e^{-y}} = \int dy - \int \frac{e^{-y}}{1+e^{-y}}\ dy= y+\ln (1+e^{-y}) + c$ (3)

The integral (3) is not 'trascendental' but may be that someone prefers to use Wolfram because 'it saves time'... well!... please observe the result supplied by Wolfram ...

Wolfram Mathematica Online Integrator

Kind regards

$\chi$ $\sigma$

7. ## Re: Be carefully using the results of Wolfram!...

Just so everyone knows, in Australia Separation of Variables isn't taught until university. The OP's question is from Year 12 Specialist Mathematics, so the method I gave is the method that is expected.

8. ## Re: Be carefully using the results of Wolfram!...

Originally Posted by Prove It
Just so everyone knows, in Australia Separation of Variables isn't taught until university. The OP's question is from Year 12 Specialist Mathematics, so the method I gave is the method that is expected.
Of course different methods can give the same result... in this case the 'Year 12 Specialist Mathematics method' gives the solution...

$x= \ln (1+e^{y}) + c$ (1)

... and the 'chisigma method' gives the solution...

$x= y + \ln (1+e^{-y}) + c = y + \ln (1+e^{-y}) + c$ (2)

... but considering the 'identity'...

$\ln (1+e^{y}) = \ln e^{y} + \ln (1+e^{-y})= y + \ln (1+e^{-y})$ (3)

... it is obvious that (1) and (2) are exactly the same expression. Scope [a little polemical may be...] of my post was to do some considerations about the 'reliability' of some 'very popular tools' ...

Kind regards

$\chi$ $\sigma$