Question

Q.1 Solve the following differential equations for y: (a) dy/dx - 4x^3 = e^2x given that y(0) = 1 (20 marks) (b) dy/dx = (5x - 3)(y + 4) given y(0) = 3 (20 marks) (c) 1/4 dy/dx = e^(2x - 5y) when y(1) = 0 (20 marks) (d) 4x dy/dx = 12x^3y^2 + 8y^2 sqrt(x) when y(2) = 1 (20 marks) Q. 2 Solve the following differential equation for r: 5 dr/d? + 3 cos(2?) = sin(4?) when r(?) = 0.95 (20 marks)

          Q.1 Solve the following differential equations for y:
(a) dy/dx - 4x^3 = e^2x given that y(0) = 1 (20 marks)
(b) dy/dx = (5x - 3)(y + 4) given y(0) = 3 (20 marks)
(c) 1/4 dy/dx = e^(2x - 5y) when y(1) = 0 (20 marks)
(d) 4x dy/dx = 12x^3y^2 + 8y^2 sqrt(x) when y(2) = 1 (20 marks)
Q. 2 Solve the following differential equation for r:
5 dr/d? + 3 cos(2?) = sin(4?) when r(?) = 0.95 (20 marks)

Q.1 Solve the following differential equations for y:
(a) dy/dx - 4x^3 = e^2x given that y(0) = 1 (20 marks)
(b) dy/dx = (5x - 3)(y + 4) given y(0) = 3 (20 marks)
(c) 1/4 dy/dx = e^(2x - 5y) when y(1) = 0 (20 marks)
(d) 4x dy/dx = 12x^3y^2 + 8y^2 sqrt(x) when y(2) = 1 (20 marks)
Q. 2 Solve the following differential equation for r:
5 dr/d? + 3 cos(2?) = sin(4?) when r(?) = 0.95 (20 marks)

Added by Mercedes E.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/06/2023

Step 1

To solve this, we first find the integrating factor, which is $e^{-\int -4 dx} = e^{4x}$. Multiply both sides of the equation by the integrating factor: $e^{4x}\frac{dy}{dx} - 4xe^{4x} = e^{5x}$ Now, notice that the left side is the derivative of Show more…

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Solve the following differential equations for y: (a) dy/dx - 4x^3 = e^2x given that y(0) = 1 (b) dy/dx = (5x - 3)(y + 4) given y(0) = 3 (c) 1/4 dy/dx = e^(2x-5y) when y(1) = 0 (d) 4x dy/dx = 12x^3y^2 + 8y^2 sqrt(x) when y(2) = 1 Q. 2 Solve the following differential equation for r: 5 dr/dθ + 3 cos(2θ) = sin(4θ) when r(π) = 0.95