Question

Show that the function y = ae^2X + be^–x is a solution of differential equation $fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y space equals space 0$ .

Solution

Here $straight y space equals space straight a space straight e to the power of 2 straight x end exponent plus straight b space straight e to the power of straight x$ ...(1)
$therefore space space space space dy over dx space equals space ae to the power of 2 straight x end exponent. space 2 space plus space straight b space straight e to the power of negative straight x end exponent left parenthesis negative 1 right parenthesis space space space space space space rightwards double arrow space space space space dy over dx space equals space 2 space straight a space straight e to the power of 2 straight x end exponent space minus space straight b space straight e to the power of negative straight x end exponent$ ...(2)
Again differentiating w.r.t. x, we get,
$fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space 4 space straight a space straight e to the power of 2 straight x end exponent plus straight b space straight e to the power of straight x$ ...(3)
L.H.S. = $fraction numerator straight d squared straight y over denominator dx squared end fraction space minus dy over dx minus 2 straight y$
$equals space 4 space straight a space straight e to the power of 2 straight x end exponent plus space straight b space straight e to the power of straight x space minus space 2 space straight a space straight e to the power of 2 straight x end exponent plus space straight b space straight e to the power of negative straight x end exponent minus 2 space straight a space straight e to the power of 2 straight x end exponent minus space 2 space straight b space straight e to the power of negative straight x end exponent space space space space left square bracket because space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis comma space left parenthesis 3 right parenthesis right square bracket$
$equals space 0 space equals space straight R. straight H. straight S.$
$therefore space space space straight y space equals space straight a space straight e to the power of 2 straight x end exponent plus straight b space straight e to the power of straight x$ is a solution of the differential equation
$fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y space equals space 0$

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Vector Algebra

Show that the function y = ae^2X + be^–x is a solution of differential equation $fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y space equals space 0$ .

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Vector Algebra

Show that the function y = ae2X + be–x is a solution of differential equation .

Some More Questions From Vector Algebra Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Show that the function y = ae^2X + be^–x is a solution of differential equation $fraction numerator straight d squared straight y over denominator dx squared end fraction minus dy over dx minus 2 straight y space equals space 0$ .