Question

sin inverse dx/dy = (x+y)​

Answer 1

The complete question is

Solve $\mathsf{sin^{-1}(\frac{dx}{dy})=x+y}$

Solution :

Now, $\mathsf{sin^{-1}(\frac{dx}{dy})=x+y}$

↬ $\mathsf{\frac{dx}{dy}=sin(x+y)}$ ...(i)

Let us take x + y = z

Differentiating both sides with respect to y, we get

$\mathsf{\frac{dx}{dy}+1=\frac{dz}{dy}}$

↬ $\mathsf{\frac{dx}{dy}=\frac{dz}{dy}-1}$

From (i) no. equation, we get

$\mathsf{\frac{dz}{dy}-1=sinz}$

↬ $\mathsf{\frac{dz}{dy}=1+sinz}$

↬ $\mathsf{\frac{dz}{1+sinz}=dy}$

↬ $\mathsf{\frac{dz}{sin^{2}\frac{z}{2}+cos^{2}\frac{z}{2}+2sin\frac{z}{2}cos\frac{z}{2}}=dy}$

since $\mathsf{sin2A =2sin\frac{A}{2}cos\frac{A}{2}}$

multiplying both the numerator and denominator of the LHS by $\mathsf{sec^{2}\frac{z}{2}}$ , we get

↬ $\mathsf{\frac{sec^{2}\frac{z}{2}dz}{tan^{2}\frac{z}{2}+1+2tanz}=dy}$

↬ $\mathsf{\frac{sec^{2}\frac{z}{2}dz}{(1+tan\frac{z}{2})^{2}}=dy}$ ...(ii)

Let us take $\mathsf{1+tan\frac{z}{2}=q}$

Taking differentials, we get

$\mathsf{sec^{2}\frac{z}{2}dz=2dq}$

From (ii), we get

$\mathsf{2\frac{dq}{q^{2}}=dy}$

On integration, we get

$\mathsf{2\int \frac{dq}{q^{2}}=\int dy}$

↬ $\mathsf{-\frac{2}{q}=y+c}$ ,

where c is integral constant

↬ $\mathsf{-\frac{2}{1+tan\frac{z}{2}}=y+c}$

↬ $\boxed{\mathsf{-\frac{2}{1+tan(\frac{x+y}{2})}=y+c}}$ ,

which is the required solution of the given differential equation.