Question

The solution lof $\frac{dy}{dx}-xtan(y-x)=1$

• A. $ce\frac{e^{x^2}}{2}=sin(y-x)$
• B. $ce\frac{e^{x^2}}{2}=sin(y+x)$
• C. $ce\frac{e^{x^2}}{2}=sin(y-x{)}^2$
• D. $ce\frac{e^{x^2}}{2}=cos(y-x)$

Answer 1

The correct option is A $ce\frac{e^{x^2}}{2}=sin(y-x)$

Put y-x = z. Then $\frac{dy}{dx}-1=\frac{dz}{dx}\Rightarrow \frac{dy}{dx}=1+\frac{dz}{dx}$
Given $\frac{dy}{dx}-xtan(y-x)=1\Rightarrow 1+\frac{dz}{dx}-xtanz=1\Rightarrow \frac{dz}{dx}=xtanz\Rightarrow \frac{1}{tanz}dz=xdx\Rightarrow \int cotzdz=\int xdx$
$\Rightarrow log|sinz|-logc=\frac{x^2}{2}\Rightarrow log|\frac{sinz}{c}|=\frac{x^2}{2}=\frac{sinz}{c}=\frac{e^{x^2}}{2}\Rightarrow sin(y-x)=c\frac{e^{x^2}}{2}$
$\therefore$ The solution is $sin(y-x)=c\frac{e^{x^2}}{2}$, where c is arbitrary constant.