Question

Consider the differential equation

Answer 1

Answer:

$W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )=e^x$

Step-by-step explanation:

Let $y=e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y'=\frac{1}{2}e^{\frac{x}{2}}$

Differentiate with respect to $x$

$y''=\frac{1}{4}e^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (\frac{1}{4}e^{\frac{x}{2}} \right )-4\left ( \frac{1}{2}e^{\frac{x}{2}}\right )+e^{\frac{x}{2}}\\=e^{\frac{x}{2}}-2e^{\frac{x}{2}}+e^{\frac{x}{2}}\\=2e^{\frac{x}{2}}-2e^{\frac{x}{2}}\\=0$

So, $y=e^{\frac{x}{2}}$ is the solution of the given equation.

Now, let $y=xe^{\frac{x}{2}}$

Differentiate with respect to $x$

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

Differentiate with respect to $x$

$y''=\frac{1}{2}e^{\frac{x}{2}}+\frac{1}{2}e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )=e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}}$

Put values of $y,y',y''$ in $4y''-4y'+y=0$

$4y''-4y'+y=0\\4\left (e^{\frac{x}{2}}+\frac{1}{4}xe^{\frac{x}{2}} \right )-4\left ( e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right )\right )+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-2e^{\frac{x}{2}}(2+x)+xe^{\frac{x}{2}}\\=4e^{\frac{x}{2}}+xe^{\frac{x}{2}}-4e^{\frac{x}{2}}-2xe^{\frac{x}{2}}+xe^{\frac{x}{2}}\\=0$

To find: $W\left ( e^{\frac{x}{2}},xe^{\frac{x}{2}} \right )$

Solution:

Let $u=e^{\frac{x}{2}}\,,\,v=xe^{\frac{x}{2}}$

$W(u,v)=\left | \begin{matrix}u&v\\u'&v' \end{matrix} \right |\\=\left | \begin{matrix}e^{\frac{x}{2}}&xe^{\frac{x}{2}}\\\frac{1}{2}e^{\frac{x}{2}}&e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \end{matrix} \right |\\=e^{\frac{x}{2}}\left [ e^{\frac{x}{2}}\left ( 1+\frac{x}{2} \right ) \right ]-\frac{1}{2}e^{\frac{x}{2}}xe^{\frac{x}{2}}\\=e^x\left ( 1+\frac{x}{2} \right )-\frac{1}{2}xe^x\\=e^x+\frac{1}{2}xe^x-\frac{1}{2}xe^x\\=e^x$