Question

(x^3 - 1)y''(x) + (x^2)y'(x) + xy(x) = 0, y(0)=2, y'(0)=1

Answer 1

Answer on Question #82283 – Math – Differential Equations

Question

$$(x ^ {\wedge} 3 - 1) y ^ {\prime \prime} (x) + (x ^ {\wedge} 2) y ^ {\prime} (x) + x y (x) = 0, y (0) = 2, y ^ {\prime} (0) = 1$$

Solution

This equation of the form:

$$x ^ {2} \left(a x ^ {n} - 1\right) y ^ {\prime \prime} (x) + x \left(a p x ^ {n} + q\right) y ^ {\prime} (x) + \left(a r x ^ {n} + s\right) y (x) = 0$$

where $n = 3$, $a = 1$, $p = 1$, $q = 0$, $r = 1$, $s = 0$. Find the roots of the quadratic equations:

\left\{ \begin{array}{l} A ^ {2} - (q + 1) A - s = 0 \\ B ^ {2} - (p - 1) B + r = 0 \end{array} \right. \to \left\{ \begin{array}{l} A ^ {2} - A = 0 \\ B ^ {2} + 1 = 0 \end{array} \right. \to A _ {1} = 0, A _ {2} = 1, B _ {1} = - i, B _ {2} = i \end{array} \right.

And define parameters $c, \alpha, \beta$ and $\gamma$:

$$c = A _ {1} = 0$$ $$\alpha = (A _ {1} + B _ {1}) n ^ {- 1} = \frac {- i}{3}$$ $$\beta = (A _ {1} + B _ {2}) n ^ {- 1} = \frac {i}{3}$$ $$\gamma = 1 + (A _ {1} - A _ {2}) n ^ {- 1} = \frac {2}{3}$$

Then the solution of the original equation has the form $y(x) = x^{c}u(ax^{n}) = u(x^{3})$ where $u = u(z)$ is the general solution of the hypergeometric equation:

$$z (z - 1) u ^ {\prime \prime} (z) + ((\alpha + \beta + 1) z - \gamma) u ^ {\prime} (z) + \alpha \beta u (z) = 0$$ $$z (z - 1) u ^ {\prime \prime} (z) + \left(z - \frac {2}{3}\right) u ^ {\prime} (z) + u (z) / 9 = 0$$

$\gamma = 2 / 3$ is not an integer, the general solution of the hypergeometric equation has the form:

$$u (z) = C _ {1} F (\alpha , \beta , \gamma ; z) + C _ {2} z ^ {1 - \gamma} F (\alpha - \gamma + 1, \beta - \gamma + 1, 2 - \gamma ; z)$$ $$y (x) = u (x ^ {3}) = C _ {1} \mathrel {_ 2} F _ {1} \left(\frac {- i}{3}, \frac {i}{3}, \frac {2}{3}; x ^ {3}\right) + C _ {2} x \mathrel {_ 2} F _ {1} \left(\frac {- i}{3} + \frac {1}{3}, \frac {i}{3} + \frac {1}{3}, \frac {4}{3}; x ^ {3}\right)$$

where ${}_{2}F_{1}$ is the hypergeometric function

$${ } _ { 2 } F _ { 1 } ( l , m , k ; t ) = \sum _ { n = 0 } ^ { \infty } \frac { ( l ) _ { n } ( m ) _ { n } } { ( k ) _ { n } } \frac { t ^ { n } } { n ! }$$ $$\frac {d}{d t} \left( { } _ { 2 } F _ { 1 } ( l , m , k ; t ) \right) = \frac { l m } { k } { } _ { 2 } F _ { 1 } ( l + 1 , m + 1 , k + 1 ; t )$$

For $y(0)=2$, $y'(0)=1$:

$$y(0) = 2 = C_1 \ {}_{2}F_1\left(\frac{-i}{3}, \frac{i}{3}, \frac{2}{3}; 0\right) + C_2 0 \ {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; 0\right) \to C_1 = 2$$ $$\begin{aligned} y'(x) &= C_1 \frac{d}{dx} \left( \ {}_{2}F_1\left(\frac{-i}{3}, \frac{i}{3}, \frac{2}{3}; x^3\right) \right) + C_2 \frac{dx}{dx} \ {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; x^3\right) \\ &\quad + C_2 x \frac{d}{dx} \left( {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; x^3\right) \right) \\ &= 2 \frac{1}{9} \frac{3}{2} {}_{2}F_1\left(\frac{-i}{3} + 1, \frac{i}{3} + 1, \frac{5}{3}; x^3\right) 3x^2 + C_2 {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; x^3\right) \\ &\quad + C_2 x \frac{2}{9} \frac{3}{4} \left( {}_{2}F_1\left(\frac{-i}{3} + \frac{4}{3}, \frac{i}{3} + \frac{4}{3}, \frac{7}{3}; x^3\right)\right) 3x^2 \to y'(0) = 1 \to C_2 = 1 \end{aligned}$$ $$y(x) = 2 \ {}_{2}F_1\left(\frac{-i}{3}, \frac{i}{3}, \frac{2}{3}; x^3\right) + x \ {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; x^3\right)$$

Answer: $y(x) = 2 \ {}_{2}F_1\left(\frac{-i}{3}, \frac{i}{3}, \frac{2}{3}; x^3\right) + x \ {}_{2}F_1\left(\frac{-i}{3} + \frac{1}{3}, \frac{i}{3} + \frac{1}{3}, \frac{4}{3}; x^3\right)$.

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