Q27RP

Question

Find a general solution to the given differential equation.

$x^{2} y'' + 2 xy' - 2 y = 6 x^{- 2} + 3 x, x > 0$

Step-by-Step Solution

Verified

Answer

The general solution to the given differential equation is;

$y = {Ax}^{- 2} + Bx - \ln |x| (x - \frac{2}{x^{2}})$

1Write the auxiliary equation of the given differential equation

The differential equation is,

$x^{2} y'' + 2 xy' - 2 y = 6 x^{- 2} + 3 x . . . . . . (1)$

Let,

$\begin{array}{rcl} x & = & e^{t} \\ dx & = & e^{t} dt \\ \frac{dt}{dx} & = & e^{- t} \\ y' & = & \frac{dy}{dx} = \frac{dy}{dt} e^{- t} \\ y'' & = & \frac{d^{2} y}{{dx}^{2}} = e^{- t} (\frac{dy}{dt} e^{- t} (- 1) + \frac{d^{2} y}{{dt}^{2}} e^{- t}) = e^{- 2 t} (\frac{d^{2} y}{{dt}^{2}} - \frac{dy}{dt}) \end{array}$

Substitute the value of $x, y$ and $y''$ in the equation (1),

$\begin{array}{rcl} x^{2} y'' + 2 xy' - 2 y & = & 6 x^{- 2} + 3 x \\ e^{2 t} [e^{- 2 t} (\frac{d^{2} y}{{dt}^{2}} - \frac{dy}{dt})] + 2 e^{t} [\frac{dy}{dt} e^{- t}] - 2 y & = & 6 x^{- 2} + 3 x \\ \frac{d^{2} y}{{dt}^{2}} + \frac{dy}{dt} - 2 y & = & 6 x^{- 2} + 3 x \\ y'' + y' - 2 y & = & 6 x^{- 2} + 3 x . . . . . . (2) \end{array}$

Write the homogeneous differential equation of the equation (1),

$y'' + y' - 2 y = 0$

The auxiliary equation for the above equation $m^{2} + m - 2 = 0$

2Find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{rcl} m^{2} + m - 2 & = & 0 \\ m^{2} + 2 m - m - 2 & = & 0 \\ m (m + 2) - 1 (m + 2) & = & 0 \\ (m - 1) (m + 2) & = & 0 \end{array}$

The roots of the auxiliary equation are $m_{1} = - 2, & m_{2} = 1$ .

The complementary solution of the given equation is $y_{c} = c_{1} e^{- 2 t} + c_{2} e^{t}$ .

3Find the particular solution

Assume, the particular solution of equation (1),

$y_{p} (t) = v_{1} \frac{1}{x^{2}} + v_{2} x . . . . . . (3)$

Now,

$\begin{array}{rcl} v_{1}' \frac{1}{x^{2}} + v_{2}' x & = & 0 \\ v_{1}' \frac{1}{x^{2}} & = & - v_{2}' x \\ v_{1}' & = & - v_{2}' x^{3} . . . . . . (4) \end{array}$

And

$\frac{- 2 v_{1}'}{x^{3}} + v_{2}' = \frac{g (x)}{a} = \frac{3 x + \frac{6}{x^{2}}}{x^{2}} \frac{- 2 v_{1}'}{x^{3}} + v_{2}' = \frac{3}{x} + \frac{6}{x^{4}}$

Substitute the value of $v_{1}'$ in the above equation,

$\begin{array}{rcl} \frac{- 2 [- v_{2}' x^{3}]}{x^{3}} + v_{2}' & = & \frac{3}{x} + \frac{6}{x^{4}} \\ 3 v_{2}' & = & \frac{3}{x} + \frac{6}{x^{4}} \\ v_{2}' & = & \frac{1}{x} + \frac{2}{x^{4}} \\ \int v_{2}' & = & \int \frac{1}{x} + \frac{2}{x^{4}} \\ v_{2} & = & \ln |x| - \frac{2}{3 x^{3}} \end{array}$

Substitute the value of $v_{2}'$ in the equation (4),

$\begin{array}{rcl} v_{1}' & = & - v_{2}' x^{3} \\ v_{1}' & = & - [\frac{1}{x} + \frac{2}{x^{4}}] x^{3} \\ \int v_{1}' & = & - \int x^{2} + \frac{2}{x} \\ v_{1} & = & - \frac{x^{3}}{3} - 2 \ln |x| \end{array}$

Therefore, the particular solution of equation (1),

$d y_{p} (t) = v_{1} \frac{1}{x^{2}} + v_{2} x y_{p} (t) = [- \frac{x^{3}}{3} - 2 \ln |x|] \frac{1}{x^{2}} + [\ln |x| - \frac{2}{3 x^{3}}] x y_{p} (t) = - \frac{x}{3} - \frac{2 \ln |x|}{x^{2}} + xln |x| - \frac{2}{3 x^{2}}$

4Write the general solution

Therefore, the general solution is,

$y = y_{c} (t) + y_{p} (t) y = c_{1} e^{- 2 t} + c_{2} e^{t} - \frac{x}{3} - \frac{2 \ln |x|}{x^{2}} + xln |x| - \frac{2}{3 x^{2}}$

Substitute $x = e^{t}$ in the above equation,

$y = c_{1} e^{- 2 t} + c_{2} e^{t} - \frac{x}{3} - \frac{2 \ln |x|}{x^{2}} + xln |x| - \frac{2}{3 x^{2}} y = c_{1} x^{- 2} + c_{2} x - \frac{x}{3} - \frac{2}{3 x^{2}} - \ln |x| (x - \frac{2}{x^{2}}) y = {Ax}^{- 2} + Bx - \ln |x| (x - \frac{2}{x^{2}})$

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