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 :

Question

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 :

${{{e^2} + 1} \over {2{e^4}}}$

${1 \over {2e}}$

${{{e^2} - 1} \over {{e^3}}}$

${{{e^2} - 1} \over {2{e^3}}}$

Solution

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