# Thread: Is there a method for finding the general solution in this case?

1. ## Is there a method for finding the general solution in this case?

I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.

2. Originally Posted by Jukodan
I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.
Perhaps some clarification is needed for this post. I for one find your question incomprehensible.

CB

3. As for myself, i think y = x/3 is the general solution. To be more precise, particular solution...

5. Originally Posted by Jukodan
I am only given y= x/3 and am asked to find the general solution. So what I did was get the derivative up until the second order and I multiplied each so that if I were to add them up they would equal zero. The question I have is...how do you choose what to multiply the derivatives by? Because I took a quiz and the first time I did they had x^2y''+7xy'+9y=0 and the second time they had switched it to x^2y''+3xy-3y=0.
Find the differential equation whos solution is y=x/3:

y'=1/3, y(0)=0

or:

y''=0, y'(0)=1/3, y(0)=0

Will do, I'm sure there are others:

CB

6. General solution to $\displaystyle x^2y''+3xy'-3y=0$ is:

$\displaystyle y(x)=c_1x+\frac{c_2}{x^3}$

I wonder why you are given $\displaystyle y=\frac{x}{3}$ instead of $\displaystyle y=x$

7. Let the ED equation be...

$\displaystyle y^{''} + 3\cdot \frac{y^{'}}{x} - 3\cdot \frac{y}{x^{2}}=0$ (1)
... and we suppose to know a particular solution of it... $\displaystyle y=x$ in this case. A general procedure to find a second solution independent from the first is illustrated now...

If u and v are both solutions of (1) is...

$\displaystyle u^{''} + 3\cdot \frac{u^{'}}{x} - 3\cdot \frac{u}{x^{2}}=0$

$\displaystyle v^{''} + 3\cdot \frac{v^{'}}{x} - 3\cdot \frac{v}{x^{2}}=0$ (2)

Multiplying the first of (2) by v, the second by u and taking the difference we obtain...

$\displaystyle u^{''}\cdot v - v^{''}\cdot u + \frac{3}{x} \cdot (u'\cdot v - v'\cdot u)=0$ (3)

Setting $\displaystyle \varphi = u'\cdot v - v'\cdot u$ the (3) is written as...

$\displaystyle \varphi ' + \frac{3}{x}\cdot \varphi=0$ (4)

... and its silution is...

$\displaystyle \varphi = \frac{c_{1}}{x^{3}}$ (5)

Deviding both termes of (4) by $\displaystyle v^{2}$ and taking into account (5) we have...

$\displaystyle \frac{u'\cdot v - v'\cdot u}{v^{2}} = \frac {d}{dx} \frac{u}{v} = \frac{c_{1}}{x^{3}\cdot v^{2}}$ (6)

... the solution of which is...

$\displaystyle \frac{u}{v}= c_{2} + c_{1} \int \frac{dx}{x^{3}\cdot v^{2}}$ (7)

If we observe (7) is easy to see that if v is solution of (1), then...

$\displaystyle u=v \int \frac{dx}{x^{3}\cdot v^{2}}$ (8)

... is also solution of (1). In particular...

$\displaystyle v=x \rightarrow u= \frac{1}{x^{3}}$ (9)

... so that the general solution of (1) is...

$\displaystyle y= c_{1} x + \frac{c_{2}}{x^{3}}$ (10)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$