Find the equation of a curve passing through the point (0,0) and whose differential equation is $y^{'} = e^{x} s i n x$

Open in App

Solution

The given differential equation of the curve is
$y^{'} = e^{x} s i n x$ or $\frac{d y}{d x} = e^{x} s i n x$
On separating the variables, we get $d y = e^{x} s i n x d x$
On intagrating both sides, we get
$\int d y = \int e^{x} s i n x d x$ ...(i)
Let $I = \int e^{x} s i n x d x$ ...(ii)
$\Rightarrow I = s i n x \int e^{x} d x - \int (\frac{d}{d x} (s i n x) \int e^{x} d x) d x = s i n x \int e^{x} d x - \int c o s x . e^{x} d x \Rightarrow s i n x . e^{x} - (c o s x . \int e^{x} d x - \int (\frac{d}{d x} (c o s x) . \int e^{x} d x) d x) \Rightarrow I = s i n x . e^{x} - [c o s x . e^{x} - \int (- s i n x) . e^{x} d x] = s i n x . e^{x} - e^{x} . c o s x - \int e^{x} . s i n x . d x \Rightarrow I = s i n x . e^{x} - c o s x . e^{x} - 1 \Rightarrow 2 I = e^{x} (s i n x - c o s x) \Rightarrow I = \frac{e^{x} (s i n x - c o s x)}{1}$
On substituting this value in Eq. (i), we get
$y = \frac{e^{x} (s i n x - c o s x)}{2} + C . . . (i i i)$
Now, the curve passes through the point (0,0) or we can say x=0, y=0.
$∴ 0 = \frac{e^{0} (s i n 0 - c o s 0)}{2} + C$
$\Rightarrow 0 = \frac{1 (0 - 1)}{2} + C \Rightarrow C = \frac{1}{2}$
On substituting $C = \frac{1}{2}$ in Eq.(iii), we get
$y = \frac{e^{x} (s i n x - c o s x)}{2} + \frac{1}{2} \Rightarrow 2 y = e^{x} (s i n x - c o s x) + 1 \Rightarrow 2 y - 1 = e^{x} (s i n x - c o s x)$
Hence, the required equation of the curve is $2 y - 1 = e^{x} (s i n x - c o s x)$

Suggest Corrections

Similar questions

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

Find the equation of the curve passing through the point whose differential equation is,

Join BYJU'S Learning Program