Answer to Question #122601 in Differential Equations for Jse

Question

The indicated function y1(x) is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2,

y2 = y1(x) ∫ e^−∫P(x) dx / y upper 2 and lower 1 (x)

as instructed, to find a second solution y2(x).

x^2 y'' − xy' + 17y = 0; y1 = x sin(4 ln x)

y2 = ____

Answer 1

$x^2y^{\prime \prime}-xy^{\prime}+17y=0$ , $y_1=x\sin(4\ln x)$

Standard form of this equation is $y^{\prime \prime}-\frac{1}{x}y^{\prime}+\frac{17}{x^2}y=0$ .

For equation $y^{\prime \prime}+P(x)y^{\prime}+Q(x)y=0$ we have formula of second solution:

$y_2(x)=y_1(x)\int \frac{e^{-\int P(x)dx}}{y^2_1(x)}dx$ .

We have that $P(x)=-\frac{1}{x},$ $\ \ e^{ -\int P(x)dx}=e^{\int \frac{1}{x}dx }=e^{\ln x}=x$ .

$y_2(x)=x\sin(4\ln x)\int \frac{1}{x\sin ^2{(4\ln x})}dx$

$\int \frac{1}{x\sin ^2(4\ln x)}dx =\big[\ \ u=4\ln x, \ du=\frac{4}{x}dx \ \ \big]=\frac{1}{4}\int \frac{1}{\sin ^2 u}du=-\frac{\cot u}{4}+C=—\frac{\cot (4\ln x)}{4}+C$

$y_2(x)=-\frac{1}{4}x\sin (4\ln x)\cot (4\ln x).=-\frac{1}{4}x\cos(4\ln x)$

Answer: $y_2(x)= -\frac{1}{4}x\cos(4\ln x).$