For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

y=cosx +C and y'+sinx=0

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Solution

Given, y=cosx +C
On differentiating both sides w.r.t. x, we get
$\Rightarrow y^{'} = - s i n x$
But we have to verify, y'+sinx=0 .....(i)
On substituting the value of y' in Eq. (i), we get
LHS=y'+sinx=-sinx+sinx=0=RHS
Hence, y=cosx+C is a solution of the given differential equation.

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For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation. y=cosx +C and y'+sinx=0

Given, y=cosx +C On differentiating both sides w.r.t. x, we get ⇒y′=−sinx But we have to verify, y'+sinx=0 .....(i) On substituting the value of y' in Eq. (i), we get LHS=y'+sinx=-sinx+sinx=0=RHS Hence, y=cosx+C is a solution of the given differential equation.

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

y=cosx +C and y'+sinx=0

Given, y=cosx +C
On differentiating both sides w.r.t. x, we get
$\Rightarrow y^{'} = - s i n x$
But we have to verify, y'+sinx=0 .....(i)
On substituting the value of y' in Eq. (i), we get
LHS=y'+sinx=-sinx+sinx=0=RHS
Hence, y=cosx+C is a solution of the given differential equation.