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Solve an ODE by first writing it in canonical form.

Consider the homogeneous, 2nd-order ODE \\[\\simplify{{b}x}\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} + \\var{a}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} - \\simplify[fractionNumbers,zeroPower,unitFactor]{{b/c^2}x^{1-2a/b}y} = 0\\,.\\]

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a) To put this equation into symmetric form, we first divide by $\\simplify{{b}x}$ to get \\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}x^2} + \\simplify{{a}/({b}x)}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x} - \\simplify[fractionNumbers,zeroPower,unitFactor]{{1/c^2}x^{-2a/b}y} = 0\\,.\\]

We then multiply by $\\exp\\left(\\int\\simplify{{a}/({b}x)}\\,\\mathrm{d}x\\right) = \\simplify{x^({a}/{b})}$, after which the first two terms can be combined in the form

\\[\\dfrac{\\mathrm{d}}{\\mathrm{d}x}\\left[\\simplify{x^({a}/{b})}\\dfrac{\\mathrm{d}y}{\\mathrm{d}x}\\right] - \\simplify[fractionNumbers,zeroPower,unitFactor]{{1/c^2}x^{-a/b}y} = 0\\,.\\]

We therefore have $A = \\var[fractionNumbers]{a/b}$ and $B = \\var[fractionNumbers]{-1/c^2}$.

b) This change of variable is designed so that \\[\\simplify{x^({a}/{b})}\\dfrac{\\mathrm{d}}{\\mathrm{d}x} = \\dfrac{\\mathrm{d}}{\\mathrm{d}t}\\,.\\]

Hence the equation becomes simply \\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}t^2} - \\simplify[fractionNumbers,unitFactor]{{1/c^2}y} = 0\\,,\\]

so $c = \\var{c}$.

c) The equation in its canonical form has constant coefficients, and so we can solve it simply by looking for solutions of the form $y = \\exp(mt)$ where $m$ is a constant. The general solution can be written as

\\[y = C\\exp(t/\\var{c}) + D\\exp(-t/\\var{c})\\,,\\]

where $C$ and $D$ are arbitrary constants. Writing this in terms of the original variable, $x$, we have

\\[y = C\\exp\\left(\\simplify[fractionNumbers]{{b/c/(b-a)}x^{1-a/b}}\\right) + D\\exp\\left(\\simplify[fractionNumbers]{{-b/c/(b-a)}x^{1-a/b}}\\right)\\,.\\]

The solution that meets the given boundary conditions has $C = D = \\dfrac{1}{2}$, and can be written as

\\[y = \\cosh\\left(\\simplify[fractionNumbers]{{b/c/(b-a)}x^{1-a/b}}\\right)\\,.\\]

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The symmetric form for this equation is \\[\\dfrac{\\mathrm{d}}{\\mathrm{d}x}\\left[x^A\\dfrac{\\mathrm{d}y}{\\mathrm{d}x}\\right] + Bx^{-A}y = 0\\,\\] where $A$ and $B$ are constants. Enter the values of $A$ and $B$ below.

$A = $ [[0]] and $B = $ [[1]]

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Rewrite the equation in its canonical form. You should find that the required change of variable is $t = \\simplify[fractionNumbers]{{b/(b-a)}x^{1-a/b}}$, and that the canonical form is \\[\\dfrac{\\mathrm{d}^2y}{\\mathrm{d}t^2} - \\frac{1}{c^2}y = 0\\,\\] where $c$ is a positive integer. Enter the value of $c$ below.

$c = $ [[0]]

Obtain the general solution of the ODE and then find the particular solution for $y(x)$ that satisfies the boundary conditions $y=1$ and $\\dfrac{\\mathrm{d}y}{\\mathrm{d}x}=0$ at $x=0$. Type this solution into the box below

$y(x) = $ [[0]]