# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Xinyu Jing on October 31, 2019, 08:35:16 PM

Title: LEC0101 QUIZ5
Post by: Xinyu Jing on October 31, 2019, 08:35:16 PM
Given
$x^{2}y''+xy'+(x^{2}-0.25)y=3x^{3/2}sinx$,x>0;
$y_{1}(x)=x^{-1/2}sinx$, $y_{2}(x)=x^{-1/2}cosx$
Step 1
The equation is written in standard form as:
$y''+\frac{1}{x}y'+\frac{(x^{2}-0.25)}{x}y=3x^{-1/2}sinx$
$g(x)=3x^{-1/2}sinx$
Now further, Wronskian is evaluated as:
W($x^{-1/2}sinx$,$x^{-1/2}cosx$)=$\left | {x^{-1/2}sinx \qquad \qquad \qquad \qquad \qquad \qquad x^{-1/2}cosx} \right |$
$\left | \frac{-1}{2}x^{-1/2}sinx+x^{-1/2}cosx \qquad \frac{-1}{2}x^{-3/2}cosx-x^{-1/2}sinx \right |$
=$\frac{-1}{2}x^{-2}sinxcosx-x^{-1}sin^{2}x+\frac{1}{2}x^{-2}sinxcos-x^{-1}cos^{2}x$
=$-x^{-1}(sin^{2}x+cos^{2}x)$=$-x^{-1}$
Step 2
The parameters U1 and U2 are evaluated as:
$u_{t}=-\int{\frac{y_{2}g(x)}{W(y_{1},y_{2})}}dx$
$u_{t}=-\int{\frac{x^{-1/2}cosx*3x^{-1/2}sinx}{-x^{-1}}dx}$
$u_{t}=\int{3cosxsinxdx}$
$u_{t}=\frac{-3}{2}cos^{2}x$
$u_{2}=-\int{\frac{y_{1}g(x)}{W(y_{1},y_{2})}dx}$
$u_{2}=-\int{\frac{x^{-1/2}sinx*3x^{-1/2}sinx}{-x^{-1}dx}}dx$
$u_{2}=\int3sinxcosx-\frac{3}{2}x$
STEP 3
Futher,
Y(x)=$y_{1}u_{1}$+$y_{2}u_{2}$
Y(x)=$x^{-1/2}sinx*\frac{-3}{2}cos^{2}x+x^{-1/2}cosx(\frac{3}{2}sinxcosx-\frac{3}{2}x)$
Y(x)=$\frac{-3}{2}x^{-1/2}cos^{2}xsinx+\frac{3}{2}x^{-1/2}cos^{2}xsinx-\frac{3}{2}x^{1/2}cosx$
Y(x)=$\frac{-3}{2}x^{1/2}cosx$
Hence, the solution is $Y(x)=\frac{-3}{2}x^{1/2}cosx$