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Question 3 Find the general solution of this separable differential equation. dy/dx = xy^2 y = 2/x^2 - c y = -2/x^2 + c y = ce^{2x^2} y = -2/(x^2 + c) Question 4 Find the integrating factor ?(x) for the first-order linear differential equation below. (Do not actually solve the equation.) (x^2 + 1) dy/dx = -4xy + 8 ?(x) = (-4x + 8)/(x^2 + 1) ?(x) = (x^2 + 1)^2 ?(x) = x^2 + 1 ?(x) = -4x/(x^2 + 1)

          Question 3
Find the general solution of this separable differential equation.
dy/dx = xy^2
y = 2/x^2 - c
y = -2/x^2 + c
y = ce^{2x^2}
y = -2/(x^2 + c)

Question 4
Find the integrating factor ?(x) for the first-order linear differential equation below. (Do not actually solve the equation.)
(x^2 + 1) dy/dx = -4xy + 8
?(x) = (-4x + 8)/(x^2 + 1)
?(x) = (x^2 + 1)^2
?(x) = x^2 + 1
?(x) = -4x/(x^2 + 1)

Question 3
Find the general solution of this separable differential equation.
dy/dx = xy^2
y = 2/x^2 - c
y = -2/x^2 + c
y = ce^2x^2
y = -2/(x^2 + c)

Question 4
Find the integrating factor ?(x) for the first-order linear differential equation below. (Do not actually solve the equation.)
(x^2 + 1) dy/dx = -4xy + 8
?(x) = (-4x + 8)/(x^2 + 1)
?(x) = (x^2 + 1)^2
?(x) = x^2 + 1
?(x) = -4x/(x^2 + 1)

Added by Carla M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Avinash Vishwakarma

Step 1

The equation is not provided clearly, so I will assume it is: $$\frac{dy}{dt} = ry$$ To solve this, we can separate the variables: $$\frac{1}{y} dy = r dt$$ Now, we can integrate both sides: $$\int \frac{1}{y} dy = \int r dt$$ $$\ln|y| = rt + C_1$$ To Show more…

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Question 3 Find the general solution of this separable differential equation: dy/dx = c1 + c2y Question 4 Find the integrating factor u(x) for the first-order linear differential equation below: (Do not actually solve the equation)