Question

.
solve the differential equation
cos(x+y) dy =dx​

Answer 1

Answer:

The required equation is \tan (x+y)-\sec (x+y)=x+Ctan(x+y)−sec(x+y)=x+C

Step-by-step explanation:

Given : Expression \frac{dy}{dx}=\sin(x+y)

dx

dy

=sin(x+y)

To find : Solve for differential equation?

Solution :

\frac{dy}{dx}=\sin(x+y)

dx

dy

=sin(x+y) ………..(1)

Let, x+y=u

Differentiate with respect to x,

1+\frac{dy}{dx}=\frac{du}{dx}1+

dx

dy

=

dx

du

\frac{dy}{dx}=\frac{du}{dx}-1

dx

dy

=

dx

du

−1 .........(2)

From equation (1) and (2),

\frac{du}{dx}-1 =sin u

dx

du

−1=sinu

\frac{du}{dx}=sin u+1

dx

du

=sinu+1

(\frac{1}{1+\sin u})du=dx(

1+sinu

1

)du=dx

\frac{1-\sin u}{(1+\sin u)(1-\sin u)}du=dx

(1+sinu)(1−sinu)

1−sinu

du=dx

\frac{1-\sin u}{\cos^2u}du=dx

cos

2

u

1−sinu

du=dx

(\sec^2 u - \sec u.\tan u)du=dx(sec

2

u−secu.tanu)du=dx

Integrate both side,

\tan u-\sec u=x+Ctanu−secu=x+C

\tan (x+y)-\sec (x+y)=x+Ctan(x+y)−sec(x+y)=x+C

Therefore, The required equation is \tan (x+y)-\sec (x+y)=x+Ctan(x+y)−sec(x+y)=x+C