For each of the following differential equations find the particular solution satisfying the given condition
$\frac{d y}{d x} = e^{2 y} cos x$ ; when $x = \frac{π}{6}, y = 0$

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Solution

$\frac{d y}{d x} = e^{2 y} cos x \Rightarrow e^{- 2 y} d y = cos x d x$
On interpreting we get
$\frac{e^{- 2 y}}{- 2} = sin x + c$ , since $x = \frac{π}{6}$ and
$y = 0$
So, we get $c = - 1$ , so, $\frac{e^{- 2 y}}{- 2} = sin x - 1$

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For each of the following differential equations find the particular solution satisfying the given condition dydx=e2ycosx; when x=π6,y=0

dydx=e2ycosx⇒e−2ydy=cosxdxOn interpreting we gete−2y−2=sinx+c, since x=π6 and y=0So, we get c=−1, so, e−2y−2=sinx−1

For each of the following differential equations find the particular solution satisfying the given condition
$\frac{d y}{d x} = e^{2 y} cos x$ ; when $x = \frac{π}{6}, y = 0$

$\frac{d y}{d x} = e^{2 y} cos x \Rightarrow e^{- 2 y} d y = cos x d x$
On interpreting we get
$\frac{e^{- 2 y}}{- 2} = sin x + c$ , since $x = \frac{π}{6}$ and
$y = 0$
So, we get $c = - 1$ , so, $\frac{e^{- 2 y}}{- 2} = sin x - 1$