Question

\[ x^{2} \frac{d^{2} y}{d x^{2}}+5 x \frac{d y}{d x}-5 y=2  \log x \text {. } \] By cauchy's Homogeneous method.

Answer 1

Given differential equation is

x² \(\frac{d^2y}{dx^2}+5x\frac{dy}{dx}-5y=2log x\)

Let x = e^z then x\(\frac{dy}{dx}=\frac{dy}{dz}\) = Dy, where \(\frac{d}{dx}=D\)

and \(x^2\frac{d^2y}{dx^2}\) = D(D - 1)y

\(\therefore\) Given differential equation converts into

D(D - 1)y + 5Dy - 5y = 2z (\(\because\) log x = z)

⇒ (D² - D + 5D - 5)y = 2z

⇒ (D² + 4D - 5)y = 2z----(1)

It's auxilarly equation is

m² + 4m - 5 = 0

⇒ (m + 5)(m - 1) = 0

⇒ m + 5 = 0 or m - 1 = 0

⇒ m = -5 or m = 1

\(\therefore\) C.F. = C₁e^-5z + C₂e^z

& P.I. = \(\frac{1}{D^2+4D-5}2z\) 

\(=\frac{-2}5\cfrac1{1-\frac{D^2+4D}5}z\) 

 

\(=\frac{-2}5(1-\frac{D^2+4D}5)^{-1}z\)

\(=\frac{-2}5(1+\frac{D^2+4D}5+\frac{(D^2+4D)^2}{5^2}+....)z\)

(\(\because\) (1 - x)^-1 = 1 + x + x²+....)

\(=\frac{-2}5(z+\frac15D^2z+\frac45 Dz)\) 

\(=\frac{-2}5(z+0+\frac45)\) (\(\because\) Dz = \(\frac{dz}{dz}=1\))

\(=\frac{-2}5z-\frac8{25}\) 

\(\therefore\) Complete solution of differential equation(1) is

y = C.F. + P. I.

 = C₁(e^z)^-5 + C₂e^z- \(\frac25z-\frac8{25}\)

 = C₁x^-5 + C₂ x - \(\frac25\) log x - \(\frac8{25}\)

(\(\because\) e^z = x ⇒ z = log x)

 = \(\frac{C_1}{x^5}+C_2x - \frac25logx-\frac8{25}\) 

which is solution of given differential equation.