y=2e2x-e-x is a solution of the differential equation

Question

y=2e2x-e-x is a solution of the differential equation (A)  y2+y1+2y=0  (B)  y2-y1+2y=0  (C)  y2+y1=0  (D)  y2-y1-2y=0

Tardigrade Admin · Accepted Answer

Given that, y=2e2x-e-x On differentiating w.r.t. x, we get y1=4e2x+e-x Now, on again differentiating w.r.t. x, we get y2=8e2x-e-x Now, y2-y1-2y =8e2x-e-x-4e2x-e-x-4e2x+2e-x =0

Q. $y = 2 e^{2 x} - e^{- x}$ is a solution of the differential equation

$y_{2} + y_{1} + 2 y = 0$

$y_{2} - y_{1} + 2 y = 0$

$y_{2} + y_{1} = 0$

$y_{2} - y_{1} - 2 y = 0$

$y_{2} - 2 y_{1} - y = 0$

Given that, $y = 2 e^{2 x} - e^{- x}$
On differentiating w.r.t. $x,$ we get $y_{1} = 4 e^{2 x} + e^{- x}$
Now, on again differentiating w.r.t. $x,$
we get $y_{2} = 8 e^{2 x} - e^{- x}$ Now, $y_{2} - y_{1} - 2 y$
$= 8 e^{2 x} - e^{- x} - 4 e^{2 x} - e^{- x} - 4 e^{2 x} + 2 e^{- x}$
$= 0$

Q. y=2e2x−e−x is a solution of the differential equation

y2​+y1​+2y=0

y2​−y1​+2y=0

y2​+y1​=0

y2​−y1​−2y=0

y2​−2y1​−y=0

Given that, y=2e2x−e−x On differentiating w.r.t. x, we get y1​=4e2x+e−x Now, on again differentiating w.r.t. x, we get y2​=8e2x−e−x Now, y2​−y1​−2y =8e2x−e−x−4e2x−e−x−4e2x+2e−x =0