Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y=e^x+1:y"-y^{\prime }=0$

Answer 1

Let
$y=e^x+1$
Differentiating both sides w.r.t. $x,$
We get,
$y^{\prime }=(e^x+1{)}^{\prime }=e^x+0$
$y^{\prime }=e^x$
Again, Differentiating both sides w.r.t. x,
We get.
$y"=(e^x{)}^{\prime }$
$y^{\prime \prime }=e^x$
Taking LHS\\
$y^{\prime \prime }-y^{\prime }=e^x-e^x$
$y^{\prime \prime }-y^{\prime }=0$
Thus $LHS=RHS$
Hence verified.

Final answer:
Hence, the function $y=e^x+1$ is a solution of the differential equation $y^{\prime \prime }-y^{\prime }=0.$