Question

Y1=x^4 is a solutionto the ode x^2y"-7xy'+16y=0 use reduction of order to find another independant solution

Answer 1

Answer:

$y_2=x^4lnx$

Step-by-step explanation:

We are given that a differential equation

$x^2y''-7xy'+16y=0$

And one solution is $y_1=x^4$

We  have to find the other independent solution by using reduction order method

$y''-\frac{7}{x}y'+\frac{16}{x^2}y=0$

Compare with the equation

$y''+P(x)y'+Q(x)y=0$

Then we get P(x)=$-\frac{7}{x}['/tex] Q(x)=[tex]\frac{16}{x^2}$

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

$y_2=x^4\int\frac{e^{\frac{7}{x}}dx}}{x^8}dx$

$y_2=x^4\int\frac{e^{7ln x}}{x^8}dx$

$y_2=x^4\int\frac{x^7}{x^8}dx$

$e^{xlny}=y^x$

$y_2=x^4\int frac{1}{x}dx$

$y_2=x^4lnx$

Answer 2

Answer with explanation:

The given differential equation is

x²y" -7 x y' +1 6 y=0---------(1)

  Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\y=zx$

Substitution the value of y, y' and y" in equation (1)

→x²z' -7 x z+16 zx=0

→x² z' + 9 zx=0

→x (x z'+9 z)=0

→x=0 ∧ x z'+9 z=0

$x \frac{dz}{dx}+9 z=0\\\\\frac{dz}{z}=-9 \frac{dx}{x}\\\\ \text{Integrating both sides}\\\\ \log z=-9 \log x+\log K\\\\ \log z+\log x^9=\log K\\\\\log zx^9=\log K\\\\K=zx^9\\\\K=y'x^9\\\\K x^{-9}d x=dy\\\\\text{Integrating both sides}\\\\y=\frac{-K}{8x^8}+m$

is another independent solution.where m and K are constant of integration.