Form the differential Equation from the following equation

[snshusat161]

Homework Statement

y= a x³ + b x²

Homework Equations

The Attempt at a Solution

There are two arb. constants, so I differentiated it twice. I couldn't guess what next to do. I Need your help!

 

[hunt_mat]

Hmm. Looks like an Euler equation to me, these can be put into the form:
<br /> a_{1}x^{2}\frac{d^{2}y}{dx^{2}}+a_{2}x\frac{dy}{dx}+a_{3}y=0<br />
The solutions of this are of the form:
<br /> y=x^{n}<br />
So all you have to do (as you know the values for n) is relate the values of n to the solution.

 

[snshusat161]

Thanks for your quick reply! But I don't know Euler equation, nor the value of n. I even don't know what actually "n" is here. After researching a bit on internet, looking some examples from my book I can only make out that we have to differentiate, add, subtract or whatever we can do to strike out those constants. Can we solve it by the method I'm trying to stat?

 

[snshusat161]

@hunt-mat, can you explain it more clearly

 

[hunt_mat]

Okay, baby steps, the value of n, is either 2 or 3, these are the solutions of your soon to be derived differential equation
<br /> y=x^{n},\quad x\frac{dy}{dx}=nx^{n},\quad x^{2}\frac{d^{2}y}{dx^{2}}=n(n-1)x^{2}<br />
The sum of these three terms must sum to zero:
so
<br /> a_{1}x^{2}\frac{d^{2}y}{dx^{2}}+a_{2}x\frac{dy}{dx}+a_{3}y=[a_{1}n(n-1)+a_{2}n+a_{3}]x^{n}<br />
This will give the quadratic in n as follows:
<br /> a_{1}n^{2}+(a_{2}-a_{1})n+a_{3}=0<br />
But you know that the solution of the above quadratic in n should be 2 and 3, the quadratic for that is (n-2)(n-3)=n^{2}-5n+6=0, so compare that with the quadratic we derived to find the a_{1}, a_{2} and a_{3}. The differential equation you will be after will be:
<br /> a_{1}x^{2}\frac{d^{2}y}{dx^{2}}+a_{2}x\frac{dy}{dx }+a_{3}y=0<br />

 

[snshusat161]

anybody here to help me solve this question with any other way, preferably with some easy method.

 

[hunt_mat]

There is no other way really, the differential equation that you seek is an Euler equation of the form I posted, the epecific equation that you seek is:
<br /> x^{2}\frac{d^{2}y}{dx^{2}}-4x\frac{dy}{dx}+6y=0<br />
The way to solve this equation, is to look for solutions of the form y=x^{n}, you substitute this into the equation and you will obtain a quadratic in n, solve this quadratic and you will note that you have two solutions, this is a linear equation so you can multiply the solution by numbers and you can add the solutions together, this is where you obtain your original function from.