JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 18)
Let y = y(x) be the solution of the differential
equation,
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi $$) = a and $${{dy} \over {dx}}$$ at x = $$\pi $$ is b, then the ordered pair (a, b) is equal to :
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$, y > 0,y(0) = 1.
If y($$\pi $$) = a and $${{dy} \over {dx}}$$ at x = $$\pi $$ is b, then the ordered pair (a, b) is equal to :
(2, 1)
$$\left( {2,{3 \over 2}} \right)$$
(1, -1)
(1, 1)
Explanation
$${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$$
$$ \Rightarrow $$ $$\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}} $$
$$ \Rightarrow $$ $$\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c$$
$$ \Rightarrow $$ $$\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c$$
As y(0) = 1
$$ \therefore $$ $$\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c$$
$$ \Rightarrow $$ $$ \therefore $$ c = 4
$$ \therefore $$ $${\left( {y + 1} \right)\left( {2 + \sin x} \right)}$$ = 4
$$ \Rightarrow $$ y = $${{{2 - \sin x} \over {2 + \sin x}}}$$
Given y($$\pi $$) = a
$$ \therefore $$ y($$\pi $$) = $${{{2 - \sin \pi } \over {2 + \sin \pi }}}$$ = 1 = a
$${{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}$$
$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{x = \pi }} = $$ 1 = b
So, (a, b) = (1, 1)
$$ \Rightarrow $$ $$\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}} $$
$$ \Rightarrow $$ $$\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c$$
$$ \Rightarrow $$ $$\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c$$
As y(0) = 1
$$ \therefore $$ $$\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c$$
$$ \Rightarrow $$ $$ \therefore $$ c = 4
$$ \therefore $$ $${\left( {y + 1} \right)\left( {2 + \sin x} \right)}$$ = 4
$$ \Rightarrow $$ y = $${{{2 - \sin x} \over {2 + \sin x}}}$$
Given y($$\pi $$) = a
$$ \therefore $$ y($$\pi $$) = $${{{2 - \sin \pi } \over {2 + \sin \pi }}}$$ = 1 = a
$${{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}$$
$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_{x = \pi }} = $$ 1 = b
So, (a, b) = (1, 1)
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