Question

The solution of the difference has equation $\frac{dy}{dx}-\frac{\tan y}{x}=\frac{\tan y\sin y}{x^2}$

• A. $\frac{x}{siny}+lnx=c$
• B. $\frac{y}{\sin y}+lnx=c$
• C. $\sin y+x=c$
• D. $\ln x=c$

Answer 1

Answer: (A)

$\frac{x}{siny}+lnx=c$

Given that:

$\frac{dy}{dx}-\frac{\tan y}{x}=\frac{\tan y\sin y}{x^2}$

$\frac{1}{\tan y\sin y}\frac{dy}{dx}-\frac{1}{\sin yx}=\frac{1}{x^2}$

Put :$\frac{1}{\sin y}=t$

$\frac{1}{\tan y\sin y}\frac{dy}{dx}=\frac{-dt}{dx}$

so $\frac{-dt}{dx}-\frac{t}{x}=\frac{1}{x^2}$

$\frac{dt}{dx}+\frac{t}{x}=-\frac{1}{x^2}$

Since this is a linear differential equation so we want $I.F$

$I.F.=e^{\int \frac{1}{x}dx}$

$I.F.=x$

Now

$t\times I.F.=\int \frac{-1}{x^2}\times I.Fdx+constant$.

$t\times x=\int \frac{-1}{x^2}\times xdx+constant$.

$t\times x=\int \frac{-1}{x}dx+constant$.

$tx=-\ln x+constant$

$\frac{x}{\sin y}=-\ln x+constant$