JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 15)
Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$
where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$
If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :
where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$
If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :
$${{{e^2} + 1} \over {2{e^4}}}$$
$${1 \over {2e}}$$
$${{{e^2} - 1} \over {{e^3}}}$$
$${{{e^2} - 1} \over {2{e^3}}}$$
Explanation
When x $$ \in $$ [0, 1], then $${{dy} \over {dx}}$$ + 2y = 1
$$ \Rightarrow $$ y = $${1 \over 2}$$ + C1e$$-$$2x
$$ \because $$ y(0) = 0 $$ \Rightarrow $$ y(x) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$e$$-$$2x
Here, y(1) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$ e$$-$$2 = $${{{e^2} - 1} \over {2{e^2}}}$$
When $$x \notin \left[ {0,1} \right]$$, then $${{dy} \over {dx}}$$ + 2y = 0 $$ \Rightarrow $$ y = c2 e$$-$$2x
$$ \because $$ y(1) = $${{{e^2} - 1} \over 2}$$ $$ \Rightarrow $$ $${{{e^2} - 1} \over 2}$$ = c2e$$-$$2 $$ \Rightarrow $$ C2 = $${{{e^2} - 1} \over 2}$$
$$ \therefore $$ y(x) $$\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right)$$ = $${{{e^2} - 1} \over {2{e^3}}}$$
$$ \Rightarrow $$ y = $${1 \over 2}$$ + C1e$$-$$2x
$$ \because $$ y(0) = 0 $$ \Rightarrow $$ y(x) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$e$$-$$2x
Here, y(1) = $${1 \over 2}$$ $$-$$ $${1 \over 2}$$ e$$-$$2 = $${{{e^2} - 1} \over {2{e^2}}}$$
When $$x \notin \left[ {0,1} \right]$$, then $${{dy} \over {dx}}$$ + 2y = 0 $$ \Rightarrow $$ y = c2 e$$-$$2x
$$ \because $$ y(1) = $${{{e^2} - 1} \over 2}$$ $$ \Rightarrow $$ $${{{e^2} - 1} \over 2}$$ = c2e$$-$$2 $$ \Rightarrow $$ C2 = $${{{e^2} - 1} \over 2}$$
$$ \therefore $$ y(x) $$\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right)$$ = $${{{e^2} - 1} \over {2{e^3}}}$$
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