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

**akshatha** · March 7th 2009, 11:58 PM

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.

**Danny** · March 8th 2009, 06:30 AM

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 $y = c_1 x + c_2 e^x$

**Danny** · March 8th 2009, 06:51 AM

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 $W_{\phi} = c W_{\psi}.$

**akshatha** · March 8th 2009, 07:10 AM

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

**bkarpuz** · March 8th 2009, 09:40 AM

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

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

LinkBack

Thread Tools

Display

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

Similar Math Help Forum Discussions

Using Laplace Transform to solve Differential equations

How do i solve this system of differential equations?

Solve a problem using differential equations.

How do I solve these first order differential equations?

[SOLVED] Differential equations help, cannot solve

Search Tags