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    The diagram shows the slope field for the differential equation $\frac{dy}{dx} = \sin(x+y), -4 \leq x \leq 5, 0 \leq y \leq 5$ The graphs of the two solutions to the differential equation that pass through points (0, 1) and (0, 3) are shown. For the two solutions given, the local minimum points lie on the straight line L₁.

    Question
    HLPaper 1

    The diagram shows the slope field for the differential equation dydx=sin⁡(x+y),−4≤x≤5,0≤y≤5\frac{dy}{dx} = \sin(x+y), -4 \leq x \leq 5, 0 \leq y \leq 5dxdy​=sin(x+y),−4≤x≤5,0≤y≤5 The graphs of the two solutions to the differential equation that pass through points (0, 1) and (0, 3) are shown. For the two solutions given, the local minimum points lie on the straight line L₁.

    1.

    Find the equation of L₁, giving your answer in the form y = mx + c.

    [1]
    Verified
    Solution

    sin⁡(x+y)=0\sin (x+y)=0sin(x+y)=0

    A1 ⇒x+y=0\Rightarrow x+y=0⇒x+y=0 (M1) (the equation of L1L_{1}L1​ is) y=−xy=-xy=−x A1 [3 marks]

    2.

    For the two solutions given, the local maximum points lie on the straight line L₂. Find the equation of L₂.

    [1]
    Verified
    Solution

    x+y=πx+y=\pix+y=π OR y=−x+πy=-x+\piy=−x+π (M1)A1 [2 marks]

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