The problem asks this:

Suppose that the second order linear equation


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

  • $\begingroup$ I edited your question to $LaTeX$ify it properly. Remember to put your $\LaTeX$ inside of "\$" signs; thus "\$ \exp(z) \$" yields $\exp(x)$. Also, by "$x > 0$" do you mean "$x \to 0"$? If so, try writing it as "\$x \to 0 \$"! Cheers! $\endgroup$ – Robert Lewis Oct 15 '17 at 16:42
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    $\begingroup$ Is $x^2y''+(3x+2x^3)y-4y=0$ or is it $x^2y''+(3x+2x^3)y'-4y=0$? $\endgroup$ – Raffaele Oct 15 '17 at 16:47
  • $\begingroup$ @Raffaele yes you're right i edited it thankyou $\endgroup$ – Eden Ovadia Oct 15 '17 at 16:56
  • $\begingroup$ You would probably have to resort to frobenius' method to solve it, wolframalpha provides the answer to this in the form of hypergeometric functions, I'm fairly certain you were intended to do as the provided answer states. $\endgroup$ – Triatticus Oct 15 '17 at 20:59

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 $W(1)= y_1(1)y_2'(1)- y_1'(1)y_2(1)= -3(2)- (-4)(-2)= -6- 8= -14$.

  • $\begingroup$ Why are you differentiating W? the problem asks for W(x) not W'(x). $\endgroup$ – Eden Ovadia Oct 15 '17 at 17:36
  • $\begingroup$ Also because the solution is such a mess! $\endgroup$ – Raffaele Oct 15 '17 at 17:37
  • $\begingroup$ You differentiate the wronskian to obtain a differential equation for it, as you cannot find it directly from its definition $\endgroup$ – Triatticus Oct 15 '17 at 20:48
  • $\begingroup$ Eden Ovakia- I said solve that equation. That gives you W. If you do not know how to solve a "separable first order differential equation", where did you get this problem? $\endgroup$ – user247327 Oct 16 '17 at 18:57

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