Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

# First Order Differential Equation?

1. dy/dx = cos (x + y), y(0) = π/4

2. dy/dx = 2 + sqrt (y − 2x + 3)

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• Ian H
Lv 7
2 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

• rotchm
Lv 7
2 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?

Done!

• Vaman
Lv 7
2 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.