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)

3 Answers

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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?

    PS: dont go by vama's answer(s). he made terrible mistakes. 

    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.

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