Thread: y''= (2/x)y' - (2/x^2)y - 1/x^2

1. y''= (2/x)y' - (2/x^2)y - 1/x^2

Hey everyone, I need help with this past exam paper as I haven't been given any solutions to it so I am unsure how to find the solutions to these problems, so if someone could please show me a stepy by step answer

Consider the following initial value problem:

y''= (2/x)y' - (2/x^2)y - 1/x^2, y(1)=0, y'(1)=1

By observing that the first two terms on the right-hand side of the equation form a total derivative of a function, find the analytical solution of the problem.

2. Well do what the question suggests...

$\displaystyle \frac{d^2y}{dx^2} = \frac{2}{x}\,\frac{dy}{dx} - \frac{2}{x^2}\,y - \frac{1}{x^2}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2}{x}\,y\right) - \frac{1}{x^2}$.

Now integrate both sides...

3. The DE can be written as...

$x^{2} y^{''} - 2 x y^{'} + 2 y = 1$ (1)

... so that it is 'Euler's type'. The solution of the 'incomplete' DE is...

$y= c_{1} t^{\alpha_{1}} + c_{2} t^{\alpha_{2}}$ (2)

... where $\alpha_{1}$ and $\alpha_{2}$ are the solutions of the 'characteristic equation'...

$\alpha\ (\alpha-1) -2 \alpha + 2=0$ (3)

... i.e. $\alpha_{1}= -1$ , $\alpha_{2}= -2$. A solution of the 'complete DE' is $y= \frac{x^{2}}{4} + \frac{1}{2}$ so that the general solution of the 'complete DE' is...

$\displaystyle y= \frac{c_{1}} {x} + \frac{c_{2}}{x^{2}} + \frac{x^{2}}{4} + \frac{1}{2}$ (4)

The value of $c_{1}$ and $c_{2}$ can be derived from the 'initial conditions'...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
The DE can be written as...

$x^{2} y^{''} - 2 x y^{'} + 2 y = 1$ (1)

... so that it is 'Euler's type'. The solution of the 'incomplete' DE is...

$y= c_{1} t^{\alpha_{1}} + c_{2} t^{\alpha_{2}}$ (2)

... where $\alpha_{1}$ and $\alpha_{2}$ are the solutions of the 'characteristic equation'...

$\alpha\ (\alpha-1) -2 \alpha + 2=0$ (3)

... i.e. $\alpha_{1}= -1$ , $\alpha_{2}= -2$. A solution of the 'complete DE' is $y= \frac{x^{2}}{4} + \frac{1}{2}$ so that the general solution of the 'complete DE' is...

$\displaystyle y= \frac{c_{1}} {x} + \frac{c_{2}}{x^{2}} + \frac{x^{2}}{4} + \frac{1}{2}$ (4)

The value of $c_{1}$ and $c_{2}$ can be derived from the 'initial conditions'...

Kind regards

$\chi$ $\sigma$
As useful as this post is, it doesn't follow the method the OP was asked to follow...

5. Thanks a lot!!