JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 18)
Let y = y(x) be the solution of the differential equation
xdy $$-$$ ydx = $$\sqrt {({x^2} - {y^2})} dx$$, x $$ \ge $$ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e$$\pi$$, y = 0 and y = y(x) is $$\alpha$$e2$$\pi$$ + $$\beta$$, then the value of 10($$\alpha$$ + $$\beta$$) is equal to __________.
xdy $$-$$ ydx = $$\sqrt {({x^2} - {y^2})} dx$$, x $$ \ge $$ 1, with y(1) = 0. If the area bounded by the line x = 1, x = e$$\pi$$, y = 0 and y = y(x) is $$\alpha$$e2$$\pi$$ + $$\beta$$, then the value of 10($$\alpha$$ + $$\beta$$) is equal to __________.
Answer
4
Explanation
$$xdy - ydx = \sqrt {{x^2} - {y^2}} dx$$
dividing both sides by x2, we get
$${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$$
$$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$$
$$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$$
Integrating both side, we get
$$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $$
$${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$$
Given, y(1) = 0 $$ \Rightarrow $$ at x = 1, y = 0
$$ \therefore $$ $$ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$$
$$ \Rightarrow $$ C = 0
$$ \therefore $$ $${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$$
$$ \Rightarrow $$ y = x sin(ln(x))
$$ \therefore $$ Area $$ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$$
Let, lnx = t
$$ \Rightarrow $$ x = et
$$ \Rightarrow $$ dx = et dt
New lower limit, t = ln(1) = 0
and upper limit t = ln$$({e^{\pi {} }})$$ = $${\pi {} }$$
$$ \therefore $$ Area = $$\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$$
$$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$$
$$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$$
$$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$$
$$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$$
$$ = \alpha {e^{2\pi {} }} + \beta $$
$$ \therefore $$ $$\alpha = {1 \over 5},\beta = {1 \over 5}$$
So, $$10(\alpha + \beta ) = 4$$
dividing both sides by x2, we get
$${{xdy - ydx} \over {{x^2}}} = {{\sqrt {{x^2} - {y^2}} } \over {{x^2}}}dx$$
$$ \Rightarrow d\left( {{y \over x}} \right) = {1 \over x}\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} dx$$
$$ \Rightarrow {{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = {{dx} \over x}$$
Integrating both side, we get
$$ \Rightarrow \int {{{d\left( {{y \over x}} \right)} \over {\sqrt {1 - {{\left( {{y \over x}} \right)}^2}} }} = \int {{{dx} \over x}} } $$
$${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x) + C$$
Given, y(1) = 0 $$ \Rightarrow $$ at x = 1, y = 0
$$ \therefore $$ $$ \Rightarrow {\sin ^{ - 1}}(0) = \ln (1) + C$$
$$ \Rightarrow $$ C = 0
$$ \therefore $$ $${\sin ^{ - 1}}\left( {{y \over x}} \right) = \ln (x)$$
$$ \Rightarrow $$ y = x sin(ln(x))
$$ \therefore $$ Area $$ = \int_1^{{e^{\pi {} }}} {x\sin (\ln (x))} dx$$
Let, lnx = t
$$ \Rightarrow $$ x = et
$$ \Rightarrow $$ dx = et dt
New lower limit, t = ln(1) = 0
and upper limit t = ln$$({e^{\pi {} }})$$ = $${\pi {} }$$
$$ \therefore $$ Area = $$\int_0^{^{\pi {} }} {{e^t}\sin (t).{e^t}} dt$$
$$ = \int_0^{^{\pi {} }} {{e^{2t}}\sin t\,} dt$$
$$ = \left[ {{{{e^{2t}}} \over {({1^2} + {2^2})}}(2\sin t - 1\cos t)} \right]_0^{\pi {} }$$
$$ = {\left[ {{{{e^{2\pi {} }}} \over 5}(0 - ( - 1)) - {1 \over 5}( - 1)} \right]}$$
$$ = {{{e^{2\pi {} }}} \over 5} + {1 \over 5}$$
$$ = \alpha {e^{2\pi {} }} + \beta $$
$$ \therefore $$ $$\alpha = {1 \over 5},\beta = {1 \over 5}$$
So, $$10(\alpha + \beta ) = 4$$
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