Find the general solution of the given differential equation. x^2y'+x(x+7)y=e^x y=

Question

Find the general solution of the given differential equation.

x^{2} y^{'} + x (x + 7) y = e^{x}

y =

Answer 1

Consider the differential equation
$x^{2} y^{'} + x (x + 7) y = e^{x}$
On rewriting it, we get
$y^{'} + (1 + \frac{2}{x}) y = \frac{e^{x}}{x^{2}} . . .$ (1)
Standard form of the linear differential equation $\frac{d y}{d x} + P (x) y = f (x)$
Clearly equation (1) is in the standard form
Compare the differential equation (1) with standard orm, and identity
$P (x) = \frac{2 + 2}{x}$ and $f (x) = \frac{e^{x}}{x^{2}}$
The dunctions $P (x) = \frac{2 + 2}{x}$ and $f (x) = \frac{e^{x}}{x^{2}}$ are continuous on $(0, \infty)$
The integrating facrote is
$e^{\int P (x) d x} = e^{\int (1 + \frac{2}{x}) d x}$
$= e^{x + 2 \ln x}$
$e^{x} e^{{\ln x}^{2}}$
$= x^{2} e^{x}$
Multiply the standart form, with the integrating factor
$x^{2} e^{x} (y^{'} + (1 + \frac{2}{x}) y) = x^{2} e^{x} \cdot \frac{e^{x}}{x^{2}}$
$x^{2} e^{x} y^{'} + x^{2} e^{x} y + 2 x e^{x} y = e^{2 x}$
$x^{2} e^{x} \frac{d y}{d x} + x^{2} y \frac{d}{d x} (e^{x}) + e^{x} y \frac{d}{d x} (x^{2}) = e^{2 x}$
$\frac{d}{d x} (x^{2} e^{x} y) = e^{2 x}$
$x^{2} e^{x} y = \int e^{2 x} d x$
$x^{2} e^{x} y = \frac{1}{2} e^{2 x} + c$
$x^{2} e^{x} y = \frac{1}{2 x^{2}} e^{x} + c x^{- 2} e^{- x}$
Hence the general solution of the given differential equation is