Question

Solve the following ODE using the power series method:
(x+2)y"+xy'-y=0

Answer 1

Answer on Question #73082 – Math – differential Equations

Question

$$(x+2)y''+xy' - y = 0$$

Solution

$$(x+2)y''+xy' - y = 0$$

Comparing the given equation with the form $y'' + P(x)y' + Q(x)y = 0$

We get $P(x) = \frac{x}{x + 2}$, $Q(x) = \frac{-1}{x + 2}$

At $x=0$, both $P(x)$ and $Q(x)$ are analytic, hence at $x=0$ is an ordinary point.

Assume its solutions to be

$$Y = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots \dots \dots + a_nx^n + \dots \dots \dots$$

Then $Y' = a_1 + 2a_2x + 3a_3x^2 + \dots \dots \dots + na_nx^{n-1} + \dots \dots \dots$

$$Y'' = 2.1a_2 + 3.2a_3x + \dots \dots \dots + n(n-1)a_nx^{n-2} + \dots \dots \dots$$

Substituting these values in the given differential equation, we get

$$(x+2)(2.1a_2 + 3.2a_3x + \dots \dots \dots + n(n-1)a_nx^{n-2} + \dots \dots \dots) + x(a_1 + 2a_2x + 3a_3x^2 + \dots \dots \dots + na_nx^{n-1} + \dots \dots \dots) - a_0 + a_1x + a_2x^2 + a_3x^3 + \dots \dots \dots + a_nx^n + \dots \dots \dots = 0$$

Equating to zero, the various powers of $x$ as,

Coefficient of $x^0 = 0$

We get $a_3 = \frac{a_0}{4}$

Coefficient of $x^1 = 0$

We get $a_2 = \frac{-3a_0}{2}$

Coefficient of $x^2 = 0$

We get $a_4 = 0$

Coefficient of $x^3 = 0$

We get $a_{s} = \frac{-a_{0}}{20}$

Substituting these values in equation (1), we get

$$Y = a _ {0} + a _ {1} x \frac {- 3 a _ {0}}{2} x ^ {2} + \frac {a _ {0}}{4} x ^ {3} - \frac {a _ {0}}{2 0} x ^ {5} \dots \dots + \dots \dots$$

Hence

$$Y = a _ {0} \frac {- 3 a _ {0}}{2} x ^ {2} + \frac {a _ {0}}{4} x ^ {3} - \frac {a _ {0}}{2 0} x ^ {5} \dots \dots + a _ {1} x \dots + \dots \dots$$ $$Y = f (x) = a _ {0} \left(1 - \frac {3}{2} x ^ {2} + \frac {1}{4} x ^ {3} - \frac {1}{2 0} x ^ {5} \dots \dots\right) + a _ {1} x \dots + \dots \dots$$

which is required solution

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