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?

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- Jan 24th 2011, 04:25 AMMaverickUK821st 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? - Jan 24th 2011, 04:28 AMSENTINEL4
If you have a solution of the differential equation you can apply it to the equation and see if it's right.

- Jan 24th 2011, 04:50 AMMaverickUK82
I understand that, but how can you substitute (and consequently check) you solution to an equation such as:

$\displaystyle y+\frac{dy}{dx}tan(x)=7$ - Jan 24th 2011, 05:01 AMSENTINEL4
If you solve this equation for example you get $\displaystyle y=\frac{c}{sinx}+7$ where c i s a constant.

Now find that $\displaystyle y'=-c\frac{cosx}{(sinx)^2}$.

Your differential equation is $\displaystyle (tanx)y'+y=7$ and you know both y and y', apply them and you will see that it is true - Jan 24th 2011, 05:03 AMHallsofIvy
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?

- Jan 24th 2011, 05:27 AMMaverickUK82
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.

- Jan 24th 2011, 05:40 AMSENTINEL4
From the things we said before didn't you understand what to do?

- Jan 24th 2011, 06:09 AMMaverickUK82
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:

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

I thought i'd solved it with:

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

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 - Jan 24th 2011, 09:33 AMMaverickUK82
When integrating

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

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

But I can only see it to be:

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

Where am I going wrong? - Jan 24th 2011, 10:03 AMSENTINEL4
The solution of mathcad is right.

It is in the form $\displaystyle \int\frac{1}{\sqrt{t^2+a^2}}dt$ so you set $\displaystyle t=|a|sinhz$ and you can find it.

How you worked to find your solution? - Jan 24th 2011, 10:20 AMMaverickUK82
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:

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

Could this be the same thing in a different format? - Jan 24th 2011, 10:34 AMMaverickUK82
Ah ha!

I just found this on another website:

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

All is not lost? - Jan 24th 2011, 12:02 PMmr fantastic
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. - Jan 24th 2011, 12:07 PMMaverickUK82
Thank you Mr. Fantastic - I will research them this evening.:D