Please help! : Wronskian and first order differential equation

[dmp306]

Please help me!! I have been attempting this problem for days but could still not find a solution! Thank you!!

Use the definition of the Wronskian, W=y1y2'-y1'y2, together with the value
W = C((t^2+1)/t))e^t^2 and y1 = 1/t, to find the second solution y2.
Standard form of equation:
y'' + (-2t^5-3t^3+t)/(t^4+t^2)y' + (-2t^4-5t^2-1)/(t^4+t^2)y = 0

 

[DrPhil]

Please help me!! I have been attempting this problem for days but could still not find a solution! Thank you!!

Use the definition of the Wronskian, W=y1y2'-y1'y2, together with the value
W = C((t^2+1)/t))e^t^2 and y1 = 1/t, to find the second solution y2.
Standard form of equation:
y'' + (-2t^5-3t^3+t)/(t^4+t^2)y' + (-2t^4-5t^2-1)/(t^4+t^2)y = 0

Looks like you have done a "good" job of writing all the functions inline, but let me put them into LaTeX to improve readability.

\(\displaystyle W = y_1 y_2' - y_1' y_2 \)

\(\displaystyle \displaystyle W = C \left( \dfrac{t^2 + 1}{t} \right) e^{t^2} \)

\(\displaystyle y_1 = 1/t \)

\(\displaystyle y'' + \dfrac{-2t^5 - 3t^3 + t}{t^4 + t^2} y' + \dfrac{-2t^4 - 5t^2 - 1}{t^4 + t^2} y = 0 \)

Do those equations look right? The next thing I would do is plug y_1 into W to get a first-order equation for y_2:

\(\displaystyle \displaystyle W = \dfrac{1}{t} y_2' + \dfrac{1}{t^2} y_2 = C \left( \dfrac{t^2 + 1}{t} \right) e^{t^2} \)

\(\displaystyle y_2' + \dfrac{1}{t} y_2 = C (t^2 + 1) e^{t^2} \)

Does that get anywhere? Please show us your work!