Question

Solve the equation x^2×d^2y/dx^2+x×dy/dx +y = x^m , for all positive integer values of m.

Answer 1

Answer on Question #66150 – Math – Differential Equations

Question

Solve the equation

$$x ^ {2} \frac {d ^ {2} y}{d x ^ {2}} + x \frac {d y}{d x} + y = x ^ {m}$$

for all positive integer values of $m$ .

Solution

We have the differential equation (DE)

$$x ^ {2} \frac {d ^ {2} y}{d x ^ {2}} + x \frac {d y}{d x} + y = x ^ {m}$$

This is the Euler equation. Change the variable $x$ by $t$ :

$$x = e ^ {t}, \quad t = \ln x, \quad (x > 0)$$

Express the derivatives with respect to $x$ in terms of the derivatives with respect to $t$ :

$$\frac {d}{d x} = \frac {d}{d t} \frac {d t}{d x} = \frac {d \ln x}{d x} \frac {d}{d t} = \frac {1}{x} \frac {d}{d t} = e ^ {- t} \frac {d}{d t}$$ $$\frac {d ^ {2}}{d x ^ {2}} = \frac {d}{d x} \left(\frac {d}{d x}\right) = e ^ {- t} \frac {d}{d t} \left(e ^ {- t} \frac {d}{d t}\right) = e ^ {- 2 t} \frac {d ^ {2}}{d t ^ {2}} - e ^ {- 2 t} \frac {d}{d t}$$

Substituting in the equation we get

$$e ^ {2 t} \left(e ^ {- 2 t} \frac {d ^ {2} y}{d t ^ {2}} - e ^ {- 2 t} \frac {d y}{d t}\right) + e ^ {t} e ^ {- t} \frac {d y}{d t} + y = e ^ {m t}$$ $$\frac {d ^ {2} y}{d t ^ {2}} + y = e ^ {m t} \quad (1)$$

To solve a nonhomogeneous linear differential equation (1) we must do the following steps:

1) find the complementary function $y_{c}$ that is the general solution of the associated homogeneous DE

$$\frac {d ^ {2} y}{d t ^ {2}} + y = 0$$

2) find any particular solution $y_{p}$ of the nonhomogeneous equation

$$\frac {d ^ {2} y}{d t ^ {2}} + y = e ^ {m t}$$

The solution of the associated homogeneous DE is

$$y _ {c} = C _ {1} \cos t + C _ {2} \sin t,$$

where $C_1$ and $C_2$ are arbitrary real constants.

Assume for the particular solution

$$y _ {p} = A e ^ {m t}$$

Substituting $y_{p}$ into the given differential equation (1), we get

$$\frac {d ^ {2} y}{d t ^ {2}} + y = A m ^ {2} e ^ {m t} + A e ^ {m t} = e ^ {m t},$$

that is,

$$A m ^ {2} + A = 1,$$

hence

$$A = \frac {1}{m ^ {2} + 1}.$$

Then

$$y _ {p} = \frac {1}{m ^ {2} + 1} e ^ {m t}.$$

The general solution of the nonhomogeneous equation is

$$y = C _ {1} \cos t + C _ {2} \sin t + \frac {1}{m ^ {2} + 1} e ^ {m t}$$

Substituting $t = \ln x$ we get

$$y = C _ {1} \cos (\ln x) + C _ {2} \sin (\ln x) + \frac {1}{m ^ {2} + 1} x ^ {m}$$

Answer:

$$y = C _ {1} \cos (\ln x) + C _ {2} \sin (\ln x) + \frac {1}{m ^ {2} + 1} x ^ {m}$$

Answer provided by https://www.AssignmentExpert.com