If the solution y(x) of the given differential equation (e^y+1) xdx+e^y xdy=0 passes throu…

Question

If the solution $y(x)$ of the given differential equation $({e}^{y}+1)\cos xdx+{e}^{y}\sin xdy=0$ passes through the point $(\frac{\pi }{2},0)$, then the value of ${e}^{y(\frac{\pi }{6})}$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Given the differential equation

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

we aim to find the value of ${e}^{y(\frac{\pi }{6})}$ given that the solution $y(x)$ passes through the point $(\frac{\pi }{2},0)$.

First, we recognize that the differential equation can be rearranged as:

$d({e}^{y}\sin x)+\cos x\, dx=0.$

Integrating this expression, we obtain:

${e}^{y}\sin x+\sin x=C,$

where $C$ is a constant. Given that the solution passes through the point $(\frac{\pi }{2},0)$, we substitute these values into the equation to find $C$:

${e}^{0}\sin (\frac{\pi }{2})+\sin (\frac{\pi }{2})=C\Rightarrow 1+1=C\Rightarrow C=2.$

Thus, the equation simplifies to:

${e}^{y}\sin x+\sin x=2.$

We now need to determine the value of ${e}^{y(\frac{\pi }{6})}$. Substituting $x=\frac{\pi }{6}$ into the equation, we get:

$({e}^{y(\frac{\pi }{6})}+1)\cdot \sin (\frac{\pi }{6})=2.$

Since $\sin (\frac{\pi }{6})=\frac{1}{2}$, the equation becomes:

$({e}^{y(\frac{\pi }{6})}+1)\cdot \frac{1}{2}=2\Rightarrow {e}^{y(\frac{\pi }{6})}+1=4\Rightarrow {e}^{y(\frac{\pi }{6})}=3.$

Therefore, the value of ${e}^{y(\frac{\pi }{6})}$ is $3$.

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