Find the equation of a curve passing through | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given differential equation is $y^{\prime} = e^{x} \sin x$.This can be written as $\frac{dy}{dx} = e^{x} \sin x$.Integrating both sides with r

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given differential equation is $y^{\prime} = e^{x} \sin x$.This can be written as $\frac{dy}{dx} = e^{x} \sin x$.Integrating both sides with respect to $x$:$y = \int e^{x} \sin x \, dx + C$  ...........$(1)$To evaluate $I = \int e^{x} \sin x \, dx$,we use integration by parts:$I = \sin x \cdot e^{x} - \int \cos x \cdot e^{x} \, dx$$I = e^{x} \sin x - [\cos x \cdot e^{x} - \int(-\sin x) \cdot e^{x} \, dx]$$I = e^{x} \sin x - e^{x} \cos x - I$$2I = e^{x}(\sin x - \cos x)$$I = \frac{e^{x}(\sin x - \cos x)}{2}$Substituting this into equation $(1)$:$y = \frac{e^{x}(\sin x - \cos x)}{2} + C$  ...........$(2)$Since the curve passes through $(0,0)$,we substitute $x=0$ and $y=0$:$0 = \frac{e^{0}(\sin 0 - \cos 0)}{2} + C$$0 = \frac{1(0 - 1)}{2} + C$$0 = -\frac{1}{2} + C \Rightarrow C = \frac{1}{2}$Substituting $C = \frac{1}{2}$ into equation $(2)$:$y = \frac{e^{x}(\sin x - \cos x)}{2} + \frac{1}{2}$$2y = e^{x}(\sin x - \cos x) + 1$$2y - 1 = e^{x}(\sin x - \cos x)$.

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

Similar Questions

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled is $\quad$ (given $\log 2=0.6912$ ) (in $years$)

$A$ normal is drawn at a point $P(x, y)$ of a curve $y=f(x)$. The normal meets the $X$-axis at $Q$. If the length $l(PQ) = k$,where $k$ is a constant,find the equation of the curve passing through $(0, k)$.

If the length of the sub-tangent at any point $P$ on a curve is proportional to the abscissa of the point $P$,then the equation of that curve is ($C$ is an arbitrary constant).

The population of a city increases at the rate of $3 \%$ per year. If the population is $p$ at time $t$,then the equation of $p$ in terms of $t$ is . . . . . . .

The slope of the tangent at any point $(x, y)$ on the curve is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point $(\sqrt{2}, e)$ is $ax + by = 1$,then $\frac{b}{a} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module