Question

DIFFERENTIAL EQUATIONS.

Solve with explanation. ​

Answer 1

Answer:

Step-by-step explanation:

dy/dx=cos(x+y)+sin(x+y).

Put x+y=t

Then, differentiate w.r.t. x,

1+dy/dx=dt/dx

Now,

dt/dx−1=cos(t)+sin(t)

dt/1+sin(t)+cos(t)=dx

Then integrate both sides,

∫dt/1+sin(t)+cos(t)=∫dx

$1+sin(t)+cos(t)=2.cos^{2}^{ \frac{t}{2}} (1+tan^{\frac{t}{2} } )$

$\frac{dt}{2.cos^{2}^{ \frac{t}{2}} (1+tan^{\frac{t}{2} } )}$

∫$\frac{sec^{2} \frac{1}{2}dt }{2(1+tan\frac{t}{2}) }$

=log(tant/2+1)

log(tan(t/2)+1)=x+c

log(tan(x+y/2)+1)=x+c  

Hope it helped you!

Answer 2

Question :-

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

Answer:-

$\red{\boxed{ log \bigg( \tan( \frac{x + y}{2} \bigg) + 1) = x + c} \: }$

Explanation :-

According to the question,

assume , (x+ y ) = z

Differentiate with respect to "x"

$1 + \frac{dy}{dx} = \frac{dz}{dx} \\ \\ \frac{dy}{dx} = \frac{dz}{dx} - 1 \\ \\ \sin(z) + \cos( z ) = \frac{dz}{dx} - 1 \\ \\ dx = \frac{dz}{1 + \sin(z) + \cos(z) } \\ \\ \: \red{ \bf{integration \: on \: both \: sides}} \\ \\ \int \: dx = \int \: \frac{dz}{1 + \sin(z) + \cos(z) } \: \\ \\ \because \: 1 + \sin(z) + \cos(z) = 2 { \cos}^{2} \frac{z}{2} \bigg(1 + \tan( \frac{z}{2} ) \bigg) \\ \\ \therefore \: \int \: \frac{dt}{ \: 2 { \cos}^{2} \frac{z}{2} \bigg(1 + \tan( \frac{z}{2} ) \bigg) \: \: } = x + c \\ \\ \int \: \frac{ { \sec }^{2} \frac{z}{2} \: dz }{2 \bigg(1 + \tan( \frac{z}{2} ) \bigg) } = x + c \\ \\ log \bigg( \tan( \frac{z}{2} ) + 1 \bigg) = x + c \\ \\ \boxed{ log \bigg( \tan( \frac{x + y}{2} \bigg) + 1) = x + c}$

Hope it helps you.