beljuA

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.
the basis of solution are ${y}_{1}={x}^{-\frac{1}{2}}$ and ${y}_{2}=x\left(\frac{3}{2}\right)$

pattererX

Expert

Given:
$4{x}^{2}-3y=0,y\left(1\right)=3,{y}^{\prime }\left(1\right)=2.5,$ the basis of solution are ${y}_{1}={x}^{-\frac{1}{2}}$ and ${y}_{2}=x\left(\frac{3}{2}\right)$
$4{x}^{2}{y}^{″}-3y=0$
Let $x={e}^{t}$
$⇒\frac{dx}{dt}={e}^{t}$
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}={e}^{-t}\frac{dy}{dt}$
$⇒{e}^{t}\frac{dy}{dx}=\frac{dy}{dt}$
$⇒x\frac{dy}{dx}=\frac{dy}{dt}$
Similarly ${x}^{2}\frac{{d}^{2}y}{d{x}^{2}}=\frac{{d}^{2}y}{d{t}^{2}}-\frac{dy}{dt}$
$4{x}^{2}{y}^{″}-3y=0$
$⇒\frac{4{d}^{2}y}{d{t}^{2}}-\frac{4dy}{dt}-3y=0$
Auxiliary equation is given by:
$4{m}^{2}-4m-3=0$
$⇒m=\frac{3}{2},\frac{-1}{2}$
$y\left(t\right)={c}_{1}{e}^{\frac{3}{2t}}+{c}_{2}{e}^{\frac{-1}{2t}}$
put
$⇒y\left(x\right)={c}_{1}{x}^{\frac{3}{2}}+{c}_{2}{x}^{-\frac{1}{2}}$
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 $\left[{x}^{\frac{3}{2}},{x}^{\frac{1}{2}}\right]$ forms a basis for the solution of $4{x}^{2}{y}^{″}-3y=0$
Now, $y\left(x\right)={c}_{1}{x}^{\frac{3}{2}}+{c}_{2}{x}^{\frac{1}{2}}$
given
$y\left(1\right)=3$
$⇒3={c}_{1}+{c}_{2}$
${y}^{\prime }\left(x\right)=\frac{3}{2}{c}_{1}{x}^{\frac{1}{2}}-\frac{1}{2}{c}_{2}{x}^{\frac{3}{2}}$
${y}^{\prime }\left(x\right)=2.5$
$⇒2.5=\frac{3}{2}{c}_{1}-\frac{1}{2}{c}_{2}$
$⇒5=3{c}_{1}-{c}_{2}$
Solving we get:

$y\left(x\right)=2{x}^{\frac{3}{2}}+{x}^{-\frac{1}{2}}$

Do you have a similar question?