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12 questions
SA-9709-P3-DE3-G01 4 marks L1 Foundation Calculator Particular solutions
Solve d y d x = 2 y x \dfrac{dy}{dx}=\dfrac{2y}{x} d x d y = x 2 y for x > 0 x>0 x > 0 , given that y = 3 y=3 y = 3 when x = 1 x=1 x = 1 . Give y y y in terms of x x x .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-Q01 1 mark L1 Foundation Calculator Harder integrals
After separating d y d x = y x 2 − 1 \dfrac{dy}{dx}=\dfrac{y}{x^2-1} d x d y = x 2 − 1 y , the x x x integral is best done by
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE-Q03 1 mark L1 Foundation Calculator Particular solutions
A particular solution differs from the general solution because it
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE-011 3 marks L2 Standard Calculator Particular solutions
Given d y d x = 3 y \dfrac{dy}{dx}=3y d x d y = 3 y and y = 2 y=2 y = 2 when x = 0 x=0 x = 0 , find the exact value of y y y when x = 1 x=1 x = 1 .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-003 4 marks L2 Standard Calculator Harder integrals
Solve d y d x = y tan x \dfrac{dy}{dx}=y\tan x d x d y = y tan x for − π 2 < x < π 2 -\tfrac\pi2<x<\tfrac\pi2 − 2 π < x < 2 π , given that y = 2 y=2 y = 2 when x = 0 x=0 x = 0 . Give y y y in terms of x x x .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-001 6 marks L3 Exam Calculator Harder integrals
Solve y d y d x = x e x y\dfrac{dy}{dx}=xe^x y d x d y = x e x , given that y = 2 y=2 y = 2 when x = 0 x=0 x = 0 and y > 0 y>0 y > 0 . Give y y y in terms of x x x .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-002 6 marks L3 Exam Calculator Harder integrals
Solve d y d x = 2 y x ( x + 2 ) \dfrac{dy}{dx}=\dfrac{2y}{x(x+2)} d x d y = x ( x + 2 ) 2 y for x > 0 x>0 x > 0 , given that y = 1 y=1 y = 1 when x = 2 x=2 x = 2 . Give y y y in terms of x x x .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-004 6 marks L3 Exam Calculator Harder integrals
Solve y d y d x = cos 2 x y\dfrac{dy}{dx}=\cos^2x y d x d y = cos 2 x , given that y = 1 y=1 y = 1 when x = 0 x=0 x = 0 and y > 0 y>0 y > 0 . Find the exact value of y y y when x = π 2 x=\tfrac\pi2 x = 2 π .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-006 5 marks L3 Exam Calculator Particular solutions
Solve ( 1 + x 2 ) d y d x = x y (1+x^2)\dfrac{dy}{dx}=xy ( 1 + x 2 ) d x d y = x y , given that y = 3 y=3 y = 3 when x = 0 x=0 x = 0 and y > 0 y>0 y > 0 . Find the value of y y y when x = 3 x=\sqrt3 x = 3 .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-005 6 marks L4 Challenging Calculator Harder integrals
Solve d y d x = e 2 y cos x \dfrac{dy}{dx}=e^{2y}\cos x d x d y = e 2 y cos x , given that y = 0 y=0 y = 0 when x = 0 x=0 x = 0 . Give y y y in terms of x x x .
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE-009 7 marks L5 Extension Calculator Harder integrals
Solve d y d x = y ( x + 1 ) ( x + 2 ) \dfrac{dy}{dx}=\dfrac{y}{(x+1)(x+2)} d x d y = ( x + 1 ) ( x + 2 ) y for x ≥ 0 x\ge0 x ≥ 0 , given that y = 1 y=1 y = 1 when x = 0 x=0 x = 0 . Give y y y in terms of x x x , and state the value that y y y approaches as x x x becomes large.
Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-DE3-007 6 marks L5 Extension Calculator Harder integrals
Solve d y d x = x ( 1 + y 2 ) \dfrac{dy}{dx}=x(1+y^2) d x d y = x ( 1 + y 2 ) , given that y = 0 y=0 y = 0 when x = 0 x=0 x = 0 . Find y y y when x = 1 x=1 x = 1 , correct to 3 significant figures, and the smallest positive value of x x x at which the solution stops existing.
Salam Academy Original Practice · Proposed mark scheme