I'm trying to do this with substitution.

z=x+y

after that the teacher lost me taking the derivitive of both sides is

=>1+dy/dx=dz/dx

How?
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2 Answers

dy/dx=cos(x+y), y(0)=PI/4

I'm trying to do this with substitution. z=x+y

after that the teacher lost me taking the derivitive of both sides is

=>1+dy/dx=dz/dx  How?

 

You have z = x + y,

now differentiate both sides, with respect to x. Then

d(z)/dx = d(x + y)/dx

dz/dx = d(x)/dx + d(y)/dx

dz/dx = dx/dx + dy/dx

dz/dx = 1 + dy/dx   (and you had 1 + dy/dx = dz/dx)

Now we substitute for z = x+y and dy/dx = cos (x + y) into the underlined equation above.

dz/dx = 1 + cos(x + y)

dz/dx = 1 + cos(z)

Using cos(2A) = 2cos^2(A) - 1,  (double angle formula)

dz/dx = 1 + (2cos^2(z/2) - 1) = 2cos^2(z/2)

dz/dx = 2cos^2(z/2)

inverting the above,

dx/dz = (1/2)sec^2(z/2)

Now integrating the above,

x = tan(z/2) + Const

z = 2arctan((x + const)

Substituting back in for z = x + y,

x + y = 2arctan(x + Const)

y(x) = -x + 2arctan(x + Const)

by Level 11 User (81.5k points)

z=x+y, so, differentiating: dz/dx=1+dy/dx. Therefore dy/dx=dz/dx-1 (dx//dx=1).

The question says dy/dx=cos(x+y).

So, substituting: dz/dx-1=cos(z). We can write this: dz-dx=cos(z)dx, and dz=dx(1+cos(z)).

So dz/(1+cos(z))=dx and the variables x and z have been separated.

∫((1/(1+cos(z))dz)=x; cos2A=2cos^2(A)-1 as a trig identity, so 2cos^2(A)=1+cos(2A). Now if 2A=z, A=z/2.

Therefore 1+cos(z)=2cos^2(z/2) and we have ∫(½sec^2(z/2)dz)=tan(z/2).

We can replace z with x+y: tan((x+y)/2)=x+C where c is the constant of integration.

When x=0, y=π/4, tan(π/8)=C and tan((x+y)/2)=x+tan(π/8). This looks like the most concise way to express the equation.

tan((x+y)/2)=tan(x/2+y/2)=(tan(x/2)+tan(y/2))/(1-tan(x/2)tan(y/2)).

by Top Rated User (804k points)

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