Answer to Question #212969 in Differential Equations for danny

Question

what do you understand by ordinary paint of the equation

a₀(x)d²y/dx²+a₁(x)dy/dx+a₂(x)y=0? hence using taylor series expansion find a series solution in powers of x for the equation(2x³-3)d²y/dx²-2xdy/dx+y=0 y(0)=-1

y¹(0)=5

Answer 1

$\text{$x$ is an ordinary point of the differential equation if $a_0 \neq 0$}\\ \text{The Taylor's series is given by:}\\ y(x) = y(x_0) + y'(x_0)(x-x_0) + y'' (x_0)\frac{(x-x_0)^2}{2!} + y''' (x_0)\frac{(x-x_0)^3}{3!} \\ \text{Where $y(x_0) = y(0) = -1$ and $y'(x_0) = y'(0) = 5$ \quad From the question we were given:}\\ (2x^3-3)y'' - 2xy' + y = 0\\ \implies y'' = \frac{2xy'-y}{2x^3-3} \implies y''(0)= \frac{2(0)(5)-(-1)}{2(0)^3 - 3} = \frac{-1}{3}\\ \text{We now compute $y'''$ using quotient rule:}\\ \implies y''' = \frac{(2x^3 - 3)(2xy''+2y'-y')-(2xy'-y)6x^2}{(2x^3-3)^2} = \frac{(2x^3 - 3)(2xy''+y')-(2xy'-y)6x^2}{(2x^3-3)^2}\\ \implies y'''(0) = \frac{(2(0)^3 - 3)(2(0)\left(\frac{-1}{3}\right) + 5) - (2(0)(5) -(-1))(6(0)^2)}{(2(0)^3 - 3)^2} = \frac{-3(5)}{9} = \frac{-5}{3}\\ \implies y'''(0)= \frac{-5}{3}\\ \text{Inputing the valuesof $y(0),y'(0),y''(0)$ and $y'''(0)$ into the Taylor's series above :}\\ y(x) = -1 + 5x -\frac{1}{6}x^2 - \frac{5}{18}x^3$