Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Find the Equation of a Curve Passing Through the Point (0, 0) and Whose Differential Equation is D Y D X = E X Sin X . - Mathematics

Sum

Find the equation of a curve passing through the point (0, 0) and whose differential equation is $\frac{dy}{dx} = e^x \sin x.$

#### Solution

We have,

$\frac{dy}{dx} = e^x \sin x$

$\Rightarrow dy = e^x \sin x dx$

Integrating both sides, we get

$\int dy = \int e^x \sin x dx$

$\Rightarrow y = I + C . . . . . . . . . . \left( 1 \right)$

$\Rightarrow I = \sin x\int e^x dx - \int\left[ \frac{d}{dx}\left( \sin x \right)\int e^x dx \right]dx$

$\Rightarrow I = \sin x e^x - \int\cos x\ e^x dx$

$\Rightarrow I = \sin x e^x - \cos x\int e^x dx + \int\left[ \frac{d}{dx}\left( \cos x \right)\int e^x dx \right]dx$

$\Rightarrow I = \sin x e^x - \cos x e^x - \int\sin x e^x dx$

$\Rightarrow I = \sin x e^x - \cos x e^x - I ...........\left[\text{From (2)} \right]$

$\Rightarrow 2I = \sin x e^x - \cos x e^x$

$\Rightarrow I = \frac{1}{2} e^x \left( \sin x - \cos x \right) . . . . . . . . . \left( 3 \right)$

From (1) and (3) we get

$\therefore y = \frac{1}{2} e^x \left( \sin x - \cos x \right) + C . . . . . . . . . \left( 4 \right)$

Now equation of the curve passes through (0, 0)

Therefore when x = 0; y = 0

Putting x = 0 and y = 0 in (4) we get

$\therefore 0 = \frac{1}{2} e^0 \left( \sin 0 - \cos 0 \right) + C$

$\Rightarrow C = \frac{1}{2}$

Substituting the value of C in (4), we get

$y = \frac{1}{2} e^x \left( \sin x - \cos x \right) + \frac{1}{2}$

$\Rightarrow 2y - 1 = e^x \left( \sin x - \cos x \right)$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 70 | Page 147