Resolved: "Check that the provided functions serve as a..."

beljuA

Answered question

2021-01-04

Check that the provided functions serve as a foundation for the solutions to the provided equation and resolve the provided initial value problem.
$4 x^{2} - 3 y = 0, y (1) = 3, y^{'} (1) = 2.5,$ the basis of solution are $y_{1} = x^{- \frac{1}{2}}$ and $y_{2} = x (\frac{3}{2})$

Answer & Explanation

pattererX

Skilled2021-01-05Added 95 answers

Given:
$4 x^{2} - 3 y = 0, y (1) = 3, y^{'} (1) = 2.5,$ the basis of solution are $y_{1} = x^{- \frac{1}{2}}$ and $y_{2} = x (\frac{3}{2})$
$4 x^{2} y^{″} - 3 y = 0$
Let $x = e^{t}$
$\Rightarrow \frac{d x}{d t} = e^{t}$
$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = e^{- t} \frac{d y}{d t}$
$\Rightarrow e^{t} \frac{d y}{d x} = \frac{d y}{d t}$
$\Rightarrow x \frac{d y}{d x} = \frac{d y}{d t}$
Similarly $x^{2} \frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d t^{2}} - \frac{d y}{d t}$
$4 x^{2} y^{″} - 3 y = 0$
$\Rightarrow \frac{4 d^{2} y}{d t^{2}} - \frac{4 d y}{d t} - 3 y = 0$
Auxiliary equation is given by:
$4 m^{2} - 4 m - 3 = 0$
$\Rightarrow m = \frac{3}{2}, \frac{- 1}{2}$
$y (t) = c_{1} e^{\frac{3}{2 t}} + c_{2} e^{\frac{- 1}{2 t}}$
put $t = \ln x (x = e^{t})$
$\Rightarrow y (x) = c_{1} x^{\frac{3}{2}} + c_{2} x^{- \frac{1}{2}}$
Consequently, every solution is a linear combination of
$x^{\frac{3}{2}}$ and $x^{\frac{1}{2}}$
also $x^{\frac{3}{2}}$ and $x^{\frac{1}{2}}$ are linearly independent so $[x^{\frac{3}{2}}, x^{\frac{1}{2}}]$ forms a basis for the solution of $4 x^{2} y^{″} - 3 y = 0$
Now, $y (x) = c_{1} x^{\frac{3}{2}} + c_{2} x^{\frac{1}{2}}$
given $y (1) = 3, y^{'} (1) = 25$
$y (1) = 3$
$\Rightarrow 3 = c_{1} + c_{2}$
$y^{'} (x) = \frac{3}{2} c_{1} x^{\frac{1}{2}} - \frac{1}{2} c_{2} x^{\frac{3}{2}}$
$y^{'} (x) = 2.5$
$\Rightarrow 2.5 = \frac{3}{2} c_{1} - \frac{1}{2} c_{2}$
$\Rightarrow 5 = 3 c_{1} - c_{2}$
Solving we get:
$c_{1} = 2, c_{2} = 1$
$y (x) = 2 x^{\frac{3}{2}} + x^{- \frac{1}{2}}$

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