# Thread: Experts in differential equations, plz solve this.....

1. ## Experts in differential equations, plz solve this.....

1)
If { Ø1(x), Ø2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
W[ ψ1(x), ψ2(x) ] = c[ Ø1, Ø2 ](x) for all a ≤ x ≤ b.
where W[] is wronskian.

2)
If one of the solution of the linear equation
(1-x) y'' + xy' - y=0,
is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.

2. Originally Posted by akshatha
1)
If { Ø1(x), Ø2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
W[ ψ1(x), ψ2(x) ] = c[ Ø1, Ø2 ](x) for all a ≤ x ≤ b.
where W[] is wronskian.

2)
If one of the solution of the linear equation
(1-x) y'' + xy' - y=0,
is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.
Here's 2) Let $y = x u$ so $y' = x u' + u$ and $y'' = x u'' + 2u'$ so your ODE becomes

$(1-x)(xu'' + 2u') + x(xu'+u) - xu = 0$ or $x(1-x)u'' +(x^2-2x+2)u' = 0$.

If you let $v = u'$ your ODE then becomes $(1-x)(xu'' + 2u') + x(xu'+u) - xu = 0$ or $x(1-x)v' +(x^2-2x+2)v = 0$. Separating gives

$\frac{dv}{v} = - \frac{x^2 -2x + 2}{x(1-x)}dx$

and upon integrating

$\ln v = x + \ln |x-1| - 2 \ln |x|$ or $v = \frac{(x-1)e^x}{x^2}$

and since $u' = v$ then

$u' = \frac{(x-1)e^x}{x^2}$

Integrating again gives

$u = \frac{e^x}{x}$

and so your second solution is $y = xu = e^x$ and the general solution $y = c_1 x + c_2 e^x$

3. Here's an answer for 1). Let us assume that your functions, $\phi_1 , \phi_2$ satisfy some ODE

$a y'' + by' + cy = 0,\;\;\; a \ne 0$

So $a \phi_1'' + b\phi_1' + c\phi_1 = 0,$ and $a \phi_2'' + b\phi_2' + c\phi_2 = 0.$ Then multiplying the first by $\phi_2$ and the second by $\phi_1$ and subtracting gives

$a\left(\phi_2 \phi_1'' - \phi_1 \phi_2'' \right) + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$ or $a\left(\phi_2 \phi_1' - \phi_1 \phi_2' \right)' + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$.

If $W_{\phi} = W(\phi_1,\phi_2)$ is the usual Wronskian, then the above equation becomes

$a W_{\phi} ' + b W_{\phi} = 0$

Similarly, for $\psi_1\; \text{and}\, \psi_2$

$a W_{\psi} ' + b W_{\psi} = 0$

Thus,

$\frac{W_{\phi} ' }{W_{\phi} } = - \frac{b}{a} = \frac{W_{\psi} ' }{W_{\psi} }$ or $\frac{W_{\phi} ' }{W_{\phi} } = \frac{W_{\psi} ' }{W_{\psi} }$

from which we obtain $W_{\phi} = c W_{\psi}.$

4. Originally Posted by danny arrigo
Here's an answer for 1). Let us assume that your functions, $\phi_1 , \phi_2$ satisfy some ODE

$a y'' + by' + cy = 0,\;\;\; a \ne 0$

So $a \phi_1'' + b\phi_1' + c\phi_1 = 0,$ and $a \phi_2'' + b\phi_2' + c\phi_2 = 0.$ Then multiplying the first by $\phi_2$ and the second by $\phi_1$ and subtracting gives

$a\left(\phi_2 \phi_1'' - \phi_1 \phi_2'' \right) + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$ or $a\left(\phi_2 \phi_1' - \phi_1 \phi_2' \right)' + b \left( \phi_2 \phi_1' - \phi_1 \phi_2' \right) = 0$.

If $W_{\phi} = W(\phi_1,\phi_2)$ is the usual Wronskian, then the above equation becomes

$a W_{\phi} ' + b W_{\phi} = 0$

Similarly, for $\psi_1\; \text{and}\, \psi_2$

$a W_{\psi} ' + b W_{\psi} = 0$

Thus,

$\frac{W_{\phi} ' }{W_{\phi} } = - \frac{b}{a} = \frac{W_{\psi} ' }{W_{\psi} }$ or $\frac{W_{\phi} ' }{W_{\phi} } = \frac{W_{\psi} ' }{W_{\psi} }$

from which we obtain $W_{\phi} = c W_{\psi}.$
thanks

5. Originally Posted by akshatha
1)
If { Ø1(x), Ø2(x) } is one set of two linearly independent solution on a ≤ x ≤ b and { ψ1(x), ψ2(x) } be another set of two linearly independent solution, then show that there exists a constant c ≠ 0 such that
W[ ψ1(x), ψ2(x) ] = c[ Ø1, Ø2 ](x) for all a ≤ x ≤ b.
where W[] is wronskian.

2)
If one of the solution of the linear equation
(1-x) y'' + xy' - y=0,
is y1(x)=x; find the second solution y2(x) such that y1(x) and y2(x) are linearly independent on some interval L.
This is a well-known fact (Abel's identity) Abel's identity - Wikipedia, the free encyclopedia