Let y=f(x) satisfies the differential equation ( sin x) d y+( cos x) y d x=ex d x with f(0)=0. Then the value of displaystyle lim x arrow 0 f(x) is

Question

Let y=f(x) satisfies the differential equation ( sin x) d y+( cos x) y d x=ex d x with f(0)=0. Then the value of displaystyle lim x arrow 0 f(x) is (A) 0 (B) 1 (C) -1 (D) Does not exist

Tardigrade Admin · Accepted Answer

( sin x) d y+( cos x) y d x=ex d x d[( sin x) ⋅ y]=ex d x Integrating y sin x=ex+c Putting f(0)=0 0=1+c ⇒ c=-1 y=(ex-1/ sin x) displaystyle lim x arrow 0 (ex-1/ sin x)= displaystyle lim x arrow 0 (ex-1/x) (x/ sin x)=1

Q. Let $y = f (x)$ satisfies the differential equation $(sin x) d y + (cos x) y d x = e^{x} d x$ with $f (0) = 0$ . Then the value of $x \to 0 lim f (x)$ is

$0$

$1$

$- 1$

Does not exist

$(sin x) d y + (cos x) y d x = e^{x} d x$
$d [(sin x) \cdot y] = e^{x} d x$
Integrating $y sin x = e^{x} + c$
Putting $f (0) = 0$
$0 = 1 + c \Rightarrow c = - 1$
$y = \frac{e ^{x} - 1}{s i n x}$
$x \to 0 lim \frac{e ^{x} - 1}{sin x} = x \to 0 lim \frac{e ^{x} - 1}{x} \frac{x}{sin x} = 1$

Q. Let y=f(x) satisfies the differential equation (sinx)dy+(cosx)ydx=exdx with f(0)=0. Then the value of x→0lim​f(x) is

0

1

−1

Does not exist

(sinx)dy+(cosx)ydx=exdx d[(sinx)⋅y]=exdx Integrating ysinx=ex+c Putting f(0)=0 0=1+c⇒c=−1 y=sinxex−1​ x→0lim​sinxex−1​=x→0lim​xex−1​sinxx​=1

Q. Let $y = f (x)$ satisfies the differential equation $(sin x) d y + (cos x) y d x = e^{x} d x$ with $f (0) = 0$ . Then the value of $x \to 0 lim f (x)$ is

$0$

$1$

$- 1$

$(sin x) d y + (cos x) y d x = e^{x} d x$
$d [(sin x) \cdot y] = e^{x} d x$
Integrating $y sin x = e^{x} + c$
Putting $f (0) = 0$
$0 = 1 + c \Rightarrow c = - 1$
$y = \frac{e ^{x} - 1}{s i n x}$
$x \to 0 lim \frac{e ^{x} - 1}{sin x} = x \to 0 lim \frac{e ^{x} - 1}{x} \frac{x}{sin x} = 1$