"Let y = y(x) satisfies the equation ${{dy} \over {dx}} - |A| = 0$, for all x 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $y(\pi ) = \pi + 2$, then the value of $y\left( {{\pi \over 2}} \right)$ is :"

Question

"Let y = y(x) satisfies the equation ${{dy} \over {dx}} - |A| = 0$, for all x > 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $y(\pi ) = \pi + 2$, then the value of $y\left( {{\pi \over 2}} \right)$ is :"

Accepted Answer

"$|A| = - {y \over x} + 2\sin x + 2$

${{dy} \over {dx}} = |A|$

${{dy} \over {dx}} = - {y \over x} + 2\sin x + 2$

${{dy} \over {dx}} + {y \over x} = 2\sin x + 2$

$I.F. = {e^{\int {{1 \over x}dx} }} = x$

$\Rightarrow yx = \int {x(2\sin x + 2)dx}$

$xy = {x^2} - 2x\cos x + 2\sin x + c$ ..... (i)

Now, x = $\pi$, y = $\pi$ + 2

Use in (i)

c = 0

Now, (i) becomes

$xy = {x^2} - 2x\cos x + 2\sin x$

put $x = \pi /2$

$${\pi \over 2}y = {\left( {{\pi \over 2}} \right)^2} - 2.{\pi \over 2}\cos {\pi \over 2} + 2\sin {\pi \over 2}$$

$y({\pi \over 2}) = {{{\pi ^2}} \over 4} + 2$"

Let y = y(x) satisfies the equation ${{dy} \over {dx}} - |A| = 0$, for all x > 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $y(\pi ) = \pi + 2$, then the value of $y\left( {{\pi \over 2}} \right)$ is :

Solution

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Let y = y(x) satisfies the equation ${{dy} \over {dx}} - |A| = 0$, for all x > 0, where $$A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right]$$. If $y(\pi ) = \pi + 2$, then the value of $y\left( {{\pi \over 2}} \right)$ is :

Solution

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