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  • Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y = e^x + 1 y'' - y' = 0

1.Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

        y = e^x + 1 \qquad :\ y'' -y'=0

Answers (1)

best_answer

Given,

y = e^x + 1

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

Again, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y' }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

\implies y'' = e^x

Substituting the values of y’ and y'' in the given differential equations,

y'' - y' =  ex - ex = 0 =  RHS.

Therefore, the given function is the solution of the corresponding differential equation.

 

Posted by

HARSH KANKARIA

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