If the solution y(x) of the given differential equation (e^y+1) \cos x ~d x+e^y \sin x ~d y=0 passes through the point (\frac\pi2, 0) , then the value of e^y(\frac\pi6) is equal to \underline\hspace2cm .

Differential Equations PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q53 with Solution

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Correct answer: 3

Given differential equation

(e^y+1)\,\cos x\,dx+e^y\sin x\,dy=0

We need the solution passing through

\left(\frac{\pi}{2},0\right)

and then find

e^{y\left(\frac{\pi}{6}\right)}.

Rewrite as a differential equation in $y$ and $x$

Divide by $dx$ :

(e^y+1)\cos x + e^y\sin x\,\frac{dy}{dx}=0

So,

e^y\sin x\,\frac{dy}{dx}=-(e^y+1)\cos x

\frac{dy}{dx}=-\frac{(e^y+1)\cos x}{e^y\sin x}

\frac{dy}{dx}=-\left(1+e^{-y}\right)\cot x

This is separable.

Separate variables

\frac{dy}{1+e^{-y}}=-\cot x\,dx

Now simplify the left side:

1+e^{-y}=\frac{e^y+1}{e^y}

Hence,

\frac{1}{1+e^{-y}}=\frac{e^y}{e^y+1}

Therefore,

\frac{e^y}{e^y+1}\,dy=-\cot x\,dx

Integrate both sides

Left side:

Let

t=e^y+1 \implies dt=e^y\,dy

So,

\int \frac{e^y}{e^y+1}\,dy=\int \frac{dt}{t}=\ln|t|= \ln(e^y+1)

Right side:

\int -\cot x\,dx=-\int \frac{\cos x}{\sin x}\,dx=-\ln|\sin x|

Thus,

\ln(e^y+1)=-\ln|\sin x|+C

\ln\big((e^y+1)|\sin x|\big)=C

So,

(e^y+1)\sin x=C

Here, near the given point $x=\frac\pi2$ , $\sin x>0$ , so we can write directly without absolute value issue.

Use the initial condition

Given that the curve passes through

x=\frac\pi2,\qquad y=0

Substitute into

(e^y+1)\sin x=C

We get

(e^0+1)\sin\frac\pi2=C

(1+1)(1)=2

So,

C=2

Hence the particular solution is

(e^y+1)\sin x=2

Find $e^{y(\pi/6)}$

x=\frac\pi6,

we have

(e^{y(\pi/6)}+1)\sin\frac\pi6=2

Since

\sin\frac\pi6=\frac12,

\left(e^{y(\pi/6)}+1\right)\cdot \frac12=2

e^{y(\pi/6)}+1=4

e^{y(\pi/6)}=3

Final answer

\boxed{3}

This matches the stored correct answer.

More from Differential Equations

View written solutionFree

Correct answer: 3

Given differential equation

(e^y+1)\,\cos x\,dx+e^y\sin x\,dy=0

We need the solution passing through

\left(\frac{\pi}{2},0\right)

and then find

e^{y\left(\frac{\pi}{6}\right)}.

Rewrite as a differential equation in $y$ and $x$

Divide by $dx$ :

(e^y+1)\cos x + e^y\sin x\,\frac{dy}{dx}=0

So,

e^y\sin x\,\frac{dy}{dx}=-(e^y+1)\cos x

\frac{dy}{dx}=-\frac{(e^y+1)\cos x}{e^y\sin x}

\frac{dy}{dx}=-\left(1+e^{-y}\right)\cot x

This is separable.

Separate variables

\frac{dy}{1+e^{-y}}=-\cot x\,dx

Now simplify the left side:

1+e^{-y}=\frac{e^y+1}{e^y}

Hence,

\frac{1}{1+e^{-y}}=\frac{e^y}{e^y+1}

Therefore,

\frac{e^y}{e^y+1}\,dy=-\cot x\,dx

Integrate both sides

Left side:

Let

t=e^y+1 \implies dt=e^y\,dy

So,

\int \frac{e^y}{e^y+1}\,dy=\int \frac{dt}{t}=\ln|t|= \ln(e^y+1)

Right side:

\int -\cot x\,dx=-\int \frac{\cos x}{\sin x}\,dx=-\ln|\sin x|

Thus,

\ln(e^y+1)=-\ln|\sin x|+C

\ln\big((e^y+1)|\sin x|\big)=C

So,

(e^y+1)\sin x=C

Here, near the given point $x=\frac\pi2$ , $\sin x>0$ , so we can write directly without absolute value issue.

Use the initial condition

Given that the curve passes through

x=\frac\pi2,\qquad y=0

Substitute into

(e^y+1)\sin x=C

We get

(e^0+1)\sin\frac\pi2=C

(1+1)(1)=2

So,

C=2

Hence the particular solution is

(e^y+1)\sin x=2

Find $e^{y(\pi/6)}$

x=\frac\pi6,

we have

(e^{y(\pi/6)}+1)\sin\frac\pi6=2

Since

\sin\frac\pi6=\frac12,

\left(e^{y(\pi/6)}+1\right)\cdot \frac12=2

e^{y(\pi/6)}+1=4

e^{y(\pi/6)}=3

Final answer

\boxed{3}

This matches the stored correct answer.

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