1st Order Differential Equations: Checking

• January 24th 2011, 04:25 AM
MaverickUK82
1st Order Differential Equations: Checking
Is there an effective method of checking a solution to a differential equation once you have solved it (or thought you have)? Using a calculator or even MathCAD.

Both implicitly and explicitly and with the use of an arbitrary constant?
• January 24th 2011, 04:28 AM
SENTINEL4
If you have a solution of the differential equation you can apply it to the equation and see if it's right.
• January 24th 2011, 04:50 AM
MaverickUK82
I understand that, but how can you substitute (and consequently check) you solution to an equation such as:

$y+\frac{dy}{dx}tan(x)=7$
• January 24th 2011, 05:01 AM
SENTINEL4
If you solve this equation for example you get $y=\frac{c}{sinx}+7$ where c i s a constant.
Now find that $y'=-c\frac{cosx}{(sinx)^2}$.
Your differential equation is $(tanx)y'+y=7$ and you know both y and y', apply them and you will see that it is true
• January 24th 2011, 05:03 AM
HallsofIvy
Quote:

Originally Posted by MaverickUK82
I understand that, but how can you substitute (and consequently check) you solution to an equation such as:

$y+\frac{dy}{dx}tan(x)=7$

That's a strange question. The answer- by doing exactly what it says! Differentiate whatever y you have (using implicit differentiation if necessary) and put y and its derivative into that equation. If the problem is that you cannot differentiate your y, what is your y?
• January 24th 2011, 05:27 AM
MaverickUK82
I've had to remove previous post because I typed the wrong solution out. Sorry guys. I've wasted you time. Mods please remove this thread. I give up.
• January 24th 2011, 05:40 AM
SENTINEL4
From the things we said before didn't you understand what to do?
• January 24th 2011, 06:09 AM
MaverickUK82
I'm sorry for giving up. I'm getting really stressed out with the maths I'm trying to learn and and some aspects of this forum, but more on topic: I used the equation above as an example - the thing I wanted to check is this:

$\sqrt{t^2+9}\frac{dy}{dt}=y^2$

I thought i'd solved it with:

$y=\frac{1}{ln(t+\sqrt{t^2+9})+C}$

It's probably wrong - but I wondered whether there was an easy way of checking whether or not it was correct before trying to solve for C to find the value of the arbitrray constant for when y(0)=1
• January 24th 2011, 09:33 AM
MaverickUK82
When integrating

$\frac{1}{\sqrt{t^2+9}}$

in math cad, it gives me $asinh(\frac{1}{3}t)$

But I can only see it to be:

$ln(x+\sqrt{x^2+3^2})+C$

Where am I going wrong?
• January 24th 2011, 10:03 AM
SENTINEL4
The solution of mathcad is right.
It is in the form $\int\frac{1}{\sqrt{t^2+a^2}}dt$ so you set $t=|a|sinhz$ and you can find it.
How you worked to find your solution?
• January 24th 2011, 10:20 AM
MaverickUK82
Hmmmm...

My course has given me a book of all the things I need to know when it comes to exam - it's like the bible - lol.

I have not came across sinh.

In the section showing the standard integrals, it gives me:

$\int\frac{1}{\sqrt{t^2+a^2}}dt=ln(x+\sqrt{x^2+a^2} )$

Could this be the same thing in a different format?
• January 24th 2011, 10:34 AM
MaverickUK82
Ah ha!

I just found this on another website:

arcsinh x = ln [x + (x2 + 1)1/2]

All is not lost?
• January 24th 2011, 12:02 PM
mr fantastic
Quote:

Originally Posted by MaverickUK82
I'm sorry for giving up. I'm getting really stressed out with the maths I'm trying to learn and and some aspects of this forum, but more on topic: I used the equation above as an example - the thing I wanted to check is this:

$\sqrt{t^2+9}\frac{dy}{dt}=y^2$

I thought i'd solved it with:

$y=\frac{1}{ln(t+\sqrt{t^2+9})+C}$

It's probably wrong - but I wondered whether there was an easy way of checking whether or not it was correct before trying to solve for C to find the value of the arbitrray constant for when y(0)=1

Quote:

Originally Posted by MaverickUK82
When integrating

$\frac{1}{\sqrt{t^2+9}}$

in math cad, it gives me $asinh(\frac{1}{3}t)$

But I can only see it to be:

$ln(x+\sqrt{x^2+3^2})+C$

Where am I going wrong?

If only you had posted the real question (quoted above) in the first place, a lot of time (and frustration on your part) might have been avoided.

As you have discovered the hard way, both answers are correct. It is not uncommon for somethng like this to happen. You should research hyperbolic functions and inverse hyperbolic functions, just to fill out some useful mathematical background you haven't been formally taught.
• January 24th 2011, 12:07 PM
MaverickUK82
Thank you Mr. Fantastic - I will research them this evening.:D