Answer with explanation:
The given differential equation is
x²y" -7 x y' +1 6 y=0---------(1)
Let, y'=z
y"=z'
[tex]\frac{dy}{dx}=z\\\\y=zx[/tex]
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
[tex]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[/tex]
is another independent solution.where m and K are constant of integration.
Answer:
[tex]y_2=x^4lnx[/tex]
Step-by-step explanation:
We are given that a differential equation
[tex]x^2y''-7xy'+16y=0[/tex]
And one solution is [tex]y_1=x^4[/tex]
We have to find the other independent solution by using reduction order method
[tex]y''-\frac{7}{x}y'+\frac{16}{x^2}y=0[/tex]
Compare with the equation
[tex]y''+P(x)y'+Q(x)y=0[/tex]
Then we get P(x)=[tex]-\frac{7}{x}['/tex] Q(x)=[tex]\frac{16}{x^2}[/tex]
[tex]y_2=y_1\int\frac{e^{-\intP(x)dx}}{y^2_1}dx[/tex]
[tex]y_2=x^4\int\frac{e^{\frac{7}{x}}dx}}{x^8}dx[/tex]
[tex]y_2=x^4\int\frac{e^{7ln x}}{x^8}dx[/tex]
[tex]y_2=x^4\int\frac{x^7}{x^8}dx[/tex]
[tex]e^{xlny}=y^x[/tex]
[tex]y_2=x^4\int frac{1}{x}dx[/tex]
[tex]y_2=x^4lnx[/tex]
