Question

Find the Wronskian of two solutions of the given differential equation without solving the equation. 9. x'y'+xy(2-y 0, Bessel's equation 10. (I-x)y"-2xy+a(a+y-0, Legendre's equation

Answer 1

SOLUTION:

Given That data find the Wronskian of two solutions of the given differential equation without solving the equation.

So

We know that for the equation

"Given That data find the Wronskian of two solutions of the given differential equation without solving the equation. So We know that for the equation y"" (t) + p (t) y (t) + q (t) y (t) = 0 is given by: W (t) = c exp (- integral p (t) dt) where a is an arbitrary constant For this equation, we can rewrite it as: y"" - (t + 2)/t y' (t + 2)/t^2 y = 0 which is of the form y"" (t) + p (t) y' (t) + q (t) y) (t) = 0 with p (t) = t + 2/t Hence, - integral p (t) dt = -integral - t + 2/t dt = integral (2/t + 1) dt That is, - integral p (t) dt = 2 in (t) + t Hence, - integral p (t) dt = in (e^t t^2) so that exp (- integral p (t) dt) = e^t t^2 Hence, the required Wronskian is W(t) = ce^t t^2 where c is an arbitrary constant We can rewrite the given equation in the form"
is given by:

where is an arbitrary constant

8) For this equation, we can rewrite it as:

which is of the form
"Given That data find the Wronskian of two solutions of the given differential equation without solving the equation. So We know that for the equation y"" (t) + p (t) y (t) + q (t) y (t) = 0 is given by: W (t) = c exp (- integral p (t) dt) where a is an arbitrary constant For this equation, we can rewrite it as: y"" - (t + 2)/t y' (t + 2)/t^2 y = 0 which is of the form y"" (t) + p (t) y' (t) + q (t) y) (t) = 0 with p (t) = t + 2/t Hence, - integral p (t) dt = -integral - t + 2/t dt = integral (2/t + 1) dt That is, - integral p (t) dt = 2 in (t) + t Hence, - integral p (t) dt = in (e^t t^2) so that exp (- integral p (t) dt) = e^t t^2 Hence, the required Wronskian is W(t) = ce^t t^2 where c is an arbitrary constant We can rewrite the given equation in the form"
with

Hence,

That is,

Hence, so that $\exp \left ( -\int p(t)\,dt \right )=e^tt^2$

Hence, the required Wronskian is where is an arbitrary constant

30) We can rewrite the given equation in the form
"Given That data find the Wronskian of two solutions of the given differential equation without solving the equation. So We know that for the equation y"" (t) + p (t) y (t) + q (t) y (t) = 0 is given by: W (t) = c exp (- integral p (t) dt) where a is an arbitrary constant For this equation, we can rewrite it as: y"" - (t + 2)/t y' (t + 2)/t^2 y = 0 which is of the form y"" (t) + p (t) y' (t) + q (t) y) (t) = 0 with p (t) = t + 2/t Hence, - integral p (t) dt = -integral - t + 2/t dt = integral (2/t + 1) dt That is, - integral p (t) dt = 2 in (t) + t Hence, - integral p (t) dt = in (e^t t^2) so that exp (- integral p (t) dt) = e^t t^2 Hence, the required Wronskian is W(t) = ce^t t^2 where c is an arbitrary constant We can rewrite the given equation in the form"
as:

$y''(t)+\tan(t)y'(t)-\frac{t}{\cos(t)}y(t)=0$ where

Hence, we have

Hence,

Therefore, the required Wronskian is

31) Again, rewriting the given equation in the form $y''(x)+p(x)y'(x)+q(x)y(x)=0$ we have

$y''(x)+\frac{y'(x)}{x}+\frac{(x^2-\nu^2)y(x)}{x^2}=0$ we have

$p(x)=\frac{1}{x}$ so that $-\int p(x)\,dx=-\int \frac{1}{x}\,dx=-\ln(x)$

Hence, we have $\exp \left ( -\int p(x)\,dx \right )=\frac{1}{x}$ and so

$W(x)=\frac{c}{x}$ is the required Wronskian

32) In this case, we have $y''-\frac{2xy'}{1-x^2}+\frac{\alpha(\alpha+1)y}{(1-x^2)}=0$ is the standard form

which has $p(x)=-\frac{2x}{1-x^2}=\frac{2x}{x^2-1}$

Therefore, $-\int p(x)\,dx=-\int \frac{2x}{x^2-1}\,dx=-\ln(x^2-1)$

Hence, $\exp \left ( -\int p(x)\,dx \right )=\frac{1}{x^2-1}$ and so the required Wronskian is

$W(x)=\frac{c}{x^2-1}$

In all the above cases, is an arbitrary constant