I feel like I'm missing a crucial step but I cannot seem to solve the following differential equation.
It is given to us that:
$$ \\ 2x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-1=0 \\ $$
where $y(1)=1$ and $y(2)=2$
How would I solve this? Using which method? I have only been taught at university the method for solving this type of equation using the complementary function method.
Any help would be really appreciated!
Considering $$\\ 2x^2\frac{d^2y}{dx^2}+4x\frac{dy}{dx}-1=0 \\$$ first reduce the order using $z=\frac{dy}{dx}$. This gives $$2x^2\frac{dz}{dx}+4xz-1=0$$ The homogneous equation is then $$x\frac{dz}{dx}+2z=0$$ which is easy to integrate (since separable). So, $$z=\frac{C}{x^2}$$ Now, variation of parameter $$2x^2\frac{dz}{dx}+4xz-1=0\implies 2 C'-1=0\implies C=\frac x2+K_1\implies z=\frac{1}{2 x}+\frac{K_1}{x^2}$$ So, $$\frac{dy}{dx}=\frac{1}{2 x}+\frac{K_1}{x^2}$$ which seems to be easy.
Note that we can rewrite the equation: $$2x^2\frac{d^2y}{dx^2}+4x\frac{dx}{dy}-1=\frac{d}{dx}\left(2x^2\frac{dx}{dy}-x\right)=0$$ Ergo, we have the following for some constant $C$: $$2x^2\frac{dy}{dx}-x=C$$ Now we can solve quite easily: \begin{align*} &2x^2\frac{dy}{dx}-x=C\\ &\frac{dy}{dx}=\frac{C}{2x^2}+\frac{1}{2x}\\ &y=\int\left[ \frac{C}{2x^2}+\frac{1}{2x}\right]dx\\ &y=-\frac{C}{2x}+\frac{1}{2}\ln\left(x\right)+B \end{align*} Letting $A=\frac{C}{2}$, we have a final answer: $$y=\frac{1}{2}\ln\left(x\right)-\frac{A}{x}+B$$
Let $x=e^{s}$, then $$\frac{dy}{ds}=\frac{dy}{dx}\frac{dx}{ds}=\frac{dy}{dx}e^{s}=x\frac{dy}{dx}$$ and $$\frac{d^{2}y}{ds^{2}} =\frac{d}{ds}\left(x\frac{dy}{dx}\right) =\frac{dx}{ds}\frac{dy}{dx}+x\frac{d}{ds}\frac{dy}{dx} =x\frac{dy}{dx}+x^{2}\frac{d^{2}y}{dx^{2}} $$
So $$x\frac{dy}{dx}=\frac{dy}{ds},\quad x^{2}\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{ds^{2}}-\frac{dy}{ds} $$ The coefficients are now constant and you can use the usual techniques to solve the new equation
$2x^{2}\frac{d^{2}y}{dx^{2}} + 4x\frac{dy}{dx} -1 = 0$
Using the substitution $x=e^{z}$
$\frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz}$
$\frac{dx}{dz} = e^{z}$
$\frac{dy}{dz} = e^{z}\frac{dy}{dx}$
$\frac{dy}{dx} = e^{-z}\frac{dy}{dz}$
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(e^{-z}\frac{dy}{dz}) = \frac{dz}{dx}\cdot \frac{d}{dz}(e^{-z}\frac{dy}{dz}) = e^{-z}(e^{-z}\frac{dy}{dz} + e^{-z}\frac{d^{2}y}{dz^{2}}) = e^{-2z}(\frac{dy}{dz} + \frac{d^{2}y}{dz^{2}})$
Then substituting into the original equation:
$2e^{2z}\cdot e^{-2z}(\frac{dy}{dz} + \frac{d^{2}y}{dz^{2}}) + 4e^{z}\cdot e^{-z}\frac{dy}{dx}-1 = 0$
$\Rightarrow 2\frac{d^{2}y}{dz^{2}} + 6\frac{dy}{dz} -1 = 0$
This is now in the form you are familiar with, which you should be able to solve. If you get stuck let me know.
