Wronskian Differential Equation.

$3y''+(6/x)y'+3e^xy = 0$ and two $y_1$, $y_2$ are two partial solutions of such that $W(y_1, y_2) \ne 0$.

(where $W(y_1, y_2) = W(x)$ is the Wronskian of $y_1$, $y_2$). If it is known that $W(1) = 2$, calculate $W(10)$. I know that i have to find a function first but i do not know how. Also i can check if they are linearly independent but how can i calculate $W(10)$. I can not find any solved examples. Please help.

The Wronskian of $y_1$ and $y_2$ is defined as

$W(y_1, y_2) = \det \begin{bmatrix} y_1 & y_2 \\ y_1' & y_2' \end{bmatrix} = y_1y_2' - y_2 y_1'; \tag 1$

we may easily find the derivative

$W' = y_1'y_2' + y_1y_2'' - y_2'y_1' - y_2y_1'' = y_1y_2'' - y_2y_1''; \tag 2$

further progress is made using the given equation

$3y''+(6/x)y'+3e^xy = 0, \tag 3$

out of which the constant factor $3$ may be divided, leaving

$y'' + (2/x)y'+ e^xy = 0, \tag 4$

which we know $y_1$ and $y_2$ solve; thus we have

$W'$ $= y_1(-((2/x)y_2' + e^xy_2)) - y_2(-((2/x)y_1' + e^xy_1)) = -(2/x)y_1y_2' - e^xy_1y_2 + (2/x)y_2y_1' + e^xy_1y_2$ $= -(2/x)y_1y_2'+ (2/x)y_2y_1' = -(2/x)(y_1y_2' - y_2y_1') =-(2/x)W; \tag 5$

once the clutter of this equation is removed we are left with

$W' = -(2/x)W, \tag 6$

which is a case of Abel's identity; the unique solution to (6) taking the value $W(1)$ at $1$ is

$W(x) = W(1)\exp \left (-\displaystyle \int_1^x (2/s)\;ds \right); \tag 7$

we may easily evaluate the integral:

$\displaystyle \int_1^x (2/s)\;ds = 2\int_1^x (1/s)\;ds = 2(\ln x - \ln 1) = 2\ln x = \ln x^2; \tag 8$

thus,

$W(x) = W(1)\exp(-\ln x^2) = W(1)\exp(\ln x^{-2}) = W(1)x^{-2}. \tag 9$

Now with

$W(1) = 2, \; x = 10, \tag{10}$

$W(x) = 2(10)^{-2} = 2(.01) = .02. \tag{11}$