The problem asks this:
Suppose that the second order linear equation
$$x^2y''+(3x+2x^3)y'-4y=0$$
has a fundamental set of solutions $\{y_1,y_2\}$ such that $y_1(1)=−3$, $y′_1(1)=2$, $y_2(1)=−2$ and $y′_2(1)=−4$
Then find explicitly the Wronskian $w(x)=y_1(x)y′_2(x)−y_2(x)y′_1(x)$
as $x>0$
I tried using $y=e^{rt}$ but I think this technique doesn't work for these types of problems. Any help at all is greatly appreciated!! Thank you
I don't understand your "I tried $y= e^{rt}$. The problem does not ask you to solve the equation! I think you are completely misunderstanding this problem.
$$W(x)= y_1'(x)y_2(x)- y_1(x)y_2'(x)$$
Differentiating with respect to $x$, \begin{aligned} W'(x) &= y_1''y_2(x)+ y_1'(x)y_2'(x)- y_1'(x)y_2'(x)- y_1(x)y_2''(x) \\ &= y_1''(x)y_2(x)- y_1(x)y''(x) \\ &= \frac{3x+ 2x^3}{x^2}(y_1(x)y_2'(x)- y_1'(x)y_2(x))\\ &= \frac{3+ 2x^2}{x}W(x) . \end{aligned}
You want to solve the equation $W'(x)= \frac{3+ 2x^2}{x}W(x)$ with the initial condition $W(1)= y_1(1)y_2'(1)- y_1'(1)y_2(1)= -3(2)- (-4)(-2)= -6- 8= -14$.