First Order Differential Equation?
1. dy/dx = cos (x + y), y(0) = π/4
2. dy/dx = 2 + sqrt (y − 2x + 3)
- Ian HLv 72 months ago
dy/dx = cos (x + y)
Let y = u - x
du/dx = cos(u) + 1
du/[cos(u) + 1] = dx
tan(u/2) = x + c
tan[(x + y)/2] = x + c
When x = 0 y = π/4
c = tan(π/8) = √(2) - 1
tan[(x + y)/2] = x + tan(π/8)
y = 2arctan[x + tan(π/8] - x
dy/dx = 2 + √(y − 2x + 3)
Let y = v + 2x so that dy/dx = dv/dx + 2
dv/dx + 2 = 2 + √(v + 3)
dv/√(v + 3) = dx
2√(v + 3) = x + c
4v + 12 = x^2 + 2cx + c^2
4(y – 2x) + 12 = x^2 + 2cx + c^2
You finish that off
- rotchmLv 72 months ago
1. Let z = x+y --> dz/dx = 1 + dy/dx. Thus dz/dx - 1 = cos(z).
Rewrite as dz/(1+cos(z)) = dx. Integrate both sides.
2. same idea.
So, what are your answers so that we may verify with you?
PS: dont go by vama's answer(s). he made terrible mistakes.
- VamanLv 72 months ago
put z=x+y, dz/dx= 1+dy/dx=1+cos z
Integrate it. z= x+sin z+c
dy/dx = 2 + sqrt (y − 2x + 3). Now put z=y-2x+3. dz/dx= ( dy/dx-2)=z^(1/2)
1/sqrt z dz/dx=1 integrate it. 2 sqrt z=x=2 sqrt(y-2x+3) I think that this is the answer.