JEE MAIN - Mathematics (2020 - 4th September Morning Slot - No. 17)
Let y = y(x) be the solution of the differential equation,
xy'- y = x2(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to :
xy'- y = x2(xcosx + sinx), x > 0. if y ($$\pi $$) = $$\pi $$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to :
$$1 + {\pi \over 2}$$
$$2 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
$$2 + {\pi \over 2}$$
$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
Explanation
$$xy' - y = {x^2}(x{\mathop{\rm cosx}\nolimits} + sinx),\,x > 0,\,y(\pi ) = \pi $$
$$y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\} $$
$$I.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}$$
$$ \therefore $$ $$y.{1 \over x} = \int {{1 \over x}.x(x\,cos\,x + sin\,x)
dx} $$
$${{y \over x}}$$ = $$\int ( x\,\cos \,x + \sin \,x)dx$$
$$ = x\,\sin \,x + C$$
$${{y \over x}}$$ = $$x\,\sin \,x + C$$
$$ \Rightarrow y = {x^2} = \sin \,x + cx$$
x = $$\pi $$, y = $$\pi $$
$$\pi $$ = $$\pi $$c $$ \Rightarrow $$ C = 1
$$y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$
$$y' = 2x\sin x + {x^2}\cos x + 1$$
y" = $$2\sin x + 2x\cos x + 2x\cos x - {x^2}\sin x$$
y" $$\left( {{\pi \over 2}} \right) = 2 - {{{\pi ^2}} \over 4} $$
$$\Rightarrow y\left( {{\pi \over 2}} \right)$$ + y" $$\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}$$
$$y' - {1 \over x}y = x\{ x\cos x\, + \,\sin x\} $$
$$I.F. = {e^{ - \int {{1 \over 2}dx} }} = {e^{ - \ln x}} = {1 \over x}$$
$$ \therefore $$ $$y.{1 \over x} = \int {{1 \over x}.x(x\,cos\,x + sin\,x)
dx} $$
$${{y \over x}}$$ = $$\int ( x\,\cos \,x + \sin \,x)dx$$
$$ = x\,\sin \,x + C$$
$${{y \over x}}$$ = $$x\,\sin \,x + C$$
$$ \Rightarrow y = {x^2} = \sin \,x + cx$$
x = $$\pi $$, y = $$\pi $$
$$\pi $$ = $$\pi $$c $$ \Rightarrow $$ C = 1
$$y = {x^2}\sin x + x \Rightarrow y\left( {{\pi \over 2}} \right) = {{{\pi ^2}} \over 4} + {\pi \over 2}$$
$$y' = 2x\sin x + {x^2}\cos x + 1$$
y" = $$2\sin x + 2x\cos x + 2x\cos x - {x^2}\sin x$$
y" $$\left( {{\pi \over 2}} \right) = 2 - {{{\pi ^2}} \over 4} $$
$$\Rightarrow y\left( {{\pi \over 2}} \right)$$ + y" $$\left( {{\pi \over 2}} \right) = 2 + {\pi \over 2}$$
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