Find a 2nd independent solution of the given equation (given one solution)?

Question

Find a second independent solution of the given equation and one solution.

I'm confused by the hint. How does this help identify a second solution given a first solution?

Answer 1

Write the differential equation in the form $ y''+P(x)y'+Q(x)y=0$

Since $x>1 \implies x \neq 1$ so

$$y''-\frac x {x-1}y'+\frac y {x-1}=0$$

$$ \text {1)} \left (\frac {y_2} {y_1}\right )'=\frac {W(y_1,y_2)} {y_1^2}$$ $$\frac {y_2'y_1-y_1'y_2}{y_1^2}=\frac {W(y_1,y_2)} {y_1^2}$$ $$\frac {y_2'e^x-e^xy_2}{e^{2x}}=\frac {W(e^x,y_2)} {e^{2x}}$$ $${y_2'e^x-e^xy_2}= {W(e^x,y_2)}$$ $${y_2'e^x-e^xy_2}=\left| \pmatrix{e^x & y_2\\ e^x & y'_2} \right|$$ $${y_2'e^x-e^xy_2}= e^xy'_2-e^xy_2$$ 2)Then use Abel's identity ... with $P(x)=-\dfrac x{x-1}$ $$W=W_0e^{-\int P(x)dx}=W_0e^{\int \frac x {x-1}}=Ke^{\int 1+\frac 1 {x-1}}=Ke^xe^{ln|x-1|}=Ke^x(x-1) \text { since x>1}$$ $$W= e^xy'_2-e^xy_2 \implies y'_2-y_2=K(x-1)$$ I let you finish...solve for $y_2$ the equation.It's a first order linear equation.

---

Another way to integrate this equation...

$$(x-1)y''-xy'+y=0$$ $$(x-1)y''-xy'\color{red}{+y'-y'}+y=0$$ $$(x-1)y''\color{blue}{-xy'+y'}+\color{green}{y-y'}=0$$ $$(x-1)y''\color{blue}{-(x-1)y'}+y-y'=0$$ $$(x-1)(y''-y')-(y'-y)=0$$ Substitute $z=y'-y$ $$(x-1)z'-z=0$$ Integrate.. $$\int \frac {dz}{z}=\int \frac {dx}{x-1}$$ $$z=K_1(x-1)$$ $$y'-y=K_1(x-1)$$ $e^{-x}$ as inetgrating factor $$(ye^{-x})'=e^{-x}K_1(x-1)$$ $$ye^{-x}=\int e^{-x}K_1(x-1)dx+K_2$$ $$y(x)=K_1e^{x}\int e^{-x}(x-1)dx+K_2e^x$$ $$\boxed{y(x)=K_1x+K_2e^x}$$