solve the Differential Equation dy/dx=cos(x+y+1)

Question

differential equation of first order

Answer 1

Let y(x)+x+1 →ø(x)

Answer 2

Y'(x)=ø'(x) -1

Answer 3

The given is transformed in to:

Answer 4

dø/dx=1+cosø

Answer 5

So we have

Answer 6

{ dø/1+cosø={ dx

Answer 7

Now we use the z- transform for the left integrand:

Answer 8

e^iø=Z

Answer 9

dø=dz/iz

Answer 10

And 1+cosø_=1+z^2 +1/2z

Answer 11

=z^2+2z+1/2z

Answer 12

=(z+1)^2/2z

Answer 13

{dø/1+cosø →"z„

Answer 14

{2z/(z+1)^2 •dz/iz=2/i

Answer 15

{ d(z+1)/(z+1)^2=2i/z+1

Answer 16

That is,

Answer 17

{dø/1+cosø=2i/1+e^iø

Answer 18

Now 2i/1+e^iø=x+c

Answer 19

→ 1+e^iø

Answer 20

=2i/x+c

Answer 21

e^iø=2i/x+c -1

Answer 22

iø= ln(2i/x+c -)

Answer 23

→ø =i ln{x+c/2i-(x+c)}

Answer 24

Y=ø-(x+1)

Answer 25

Y=-(x+1)+i ln{x+c/2i-(x+c)}

Answer 26

Therefore,

Answer 27

solve the Differential Equation dy/dx=cos(x+y+1)

let u = x + y + 1

then,

du/dx = 1 + dy/dx, i.e.

dy/dx = du/dx – 1, so

cos(u) = du/dx – 1   (using u = x + y + 1)

Rearranging,

du/(cos(u) +1) = dx, or

int du/(cos(u) + 1) = int dx

Using the trig identity cos(2X) = 2cos^2(X) – 1, the expression becomes

int du/(2cos^2(u/2) – 1 + 1) = int dx

int du/(2cos^2(u/2)) = int dx

The lhs above is a standard integral and the whole integration becomes

tan(u/2) = x + C

 (x + y + 1)/2 = arctan(x + C)

y(x) = -x – 1 + 2arctan(x + C)

Answer 28

y'=cos(x+y+1)

Let z=x+y+1 and z'=dz/dx=1+y'=1+cos(x+y+1)=1+cosz.

dz/dx=1+cosz; ∫dz/(1+cosz)=∫dx=x.

x=∫dz/(1+cosz); cosz=2cos^2(z/2)-1, so x=½∫sec^2(z/2)dz=½(2tan(z/2))=tan(z/2)+C.

Therefore x=tan((x+y+1)/2)+C (implicit equation).

CHECK

Differential of the solution: 1=½sec^2((x+y+1)/2)(1+y')=(1+y')/(2cos^2((x+y+1)/2)=

(1+y')/(1+cos(x+y+1)), and 1+y'=1+cos(x+y+1), so dy/dx=cos(x+y+1).

The check demonstrates that x=tan((x+y+1)/2)+C is a solution in implicit form.

So tan((x+y+1)/2)=x-C and x+y+1=2arctan(x-C), y=2arctan(x-C)-x-1 (explicit form).