4.10c Integrating factor: first order equations

$$\frac{dy}{dx} + y\left(1 + \frac{3}{x}\right) = \frac{1}{x^2}, \quad x > 0.$$

1. Verify that \(x^3e^x\) is an integrating factor for the differential equation. [3]
2. Find the general solution of the differential equation. [4]
3. Given that \(y = 1\) at \(x = 1\), find \(y\) at \(x = 2\). [3]

Find the general solution of the differential equation $$\frac{dy}{dx} + 2y \cot 2x = \sin x, \quad 0 < x < \frac{\pi}{2},$$ giving your answer in the form \(y = f(x)\). [7]

Find the general solution of the differential equation $$(x + 1)\frac{dy}{dx} + 2y = \frac{1}{x}, \quad x > 0.$$ giving your answer in the form \(y = f(x)\). [7]

A particle of mass \(m\) kg moves in a horizontal straight line. Its initial speed is \(u\) ms\(^{-1}\) and the only force acting on it is a variable resistance of magnitude \(mkv\) N, where \(v\) ms\(^{-1}\) is the speed of the particle after \(t\) seconds and \(k\) is a constant. Show that \(v = ue^{-kt}\). [7 marks]

At time \(t = 0\), the position vector of a particle \(P\) is \(-3j\) m. At time \(t\) seconds, the position vector of \(P\) is \(\mathbf{r}\) metres and the velocity of \(P\) is \(\mathbf{v}\) m s\(^{-1}\). Given that $$\mathbf{v} - 2\mathbf{r} = 4e^t \mathbf{j},$$ find the time when \(P\) passes through the origin. [7]

A raindrop falls vertically under gravity through a cloud. In a model of the motion the raindrop is assumed to be spherical at all times and the cloud is assumed to consist of stationary water particles. At time \(t = 0\), the raindrop is at rest and has radius \(a\). As the raindrop falls, water particles from the cloud condense onto it and the radius of the raindrop is assumed to increase at a constant rate \(\lambda\). At time \(t\) the speed of the raindrop is \(v\).

1. Show that $$\frac{dv}{dt} + \frac{3\lambda v}{(\lambda t + a)} = g.$$ [8]

1. Find the speed of the raindrop when its radius is \(3a\). [7]

A particle \(P\) moves in a plane such that its position vector \(\mathbf{r}\) metres at time \(t\) seconds \((t > 0)\) satisfies the differential equation $$\frac{d\mathbf{r}}{dt} - \frac{2}{t}\mathbf{r} = 4i$$ When \(t = 1\), the particle is at the point with position vector \((i + j)\) m. Find \(\mathbf{r}\) in terms of \(t\). [9]

A particle \(P\) moves in a plane such that its position vector \(\mathbf{r}\) metres at time \(t\) seconds \((t > 0)\) satisfies the differential equation $$\frac{d\mathbf{r}}{dt} - \frac{2}{t}\mathbf{r} = 4\mathbf{i}$$ When \(t = 1\), the particle is at the point with position vector \((\mathbf{i} + \mathbf{j})\) m. Find \(\mathbf{r}\) in terms of \(t\). [9]

A particle \(P\) moves in a plane so that its position vector, \(\mathbf{r}\) metres at time \(t\) seconds, satisfies the differential equation $$\frac{d\mathbf{r}}{dt} + \mathbf{r} = t\mathbf{i} + e^{-t}\mathbf{j}$$ When \(t = 0\) the particle is at the point with position vector \((\mathbf{i} + \mathbf{j})\) m. Find \(\mathbf{r}\) in terms of \(t\). [9]

As a hailstone falls under gravity in still air, its mass increases. At time \(t\) the mass of the hailstone is \(m\). The hailstone is modelled as a uniform sphere of radius \(r\) such that $$\frac{dr}{dt} = kr,$$ where \(k\) is a positive constant.

1. Show that \(\frac{dm}{dt} = 3km\). [2]

Assuming that there is no air resistance,

1. show that the speed \(v\) of the hailstone at time \(t\) satisfies $$\frac{dv}{dt} = g - 3kv.$$ [4]

Given that the speed of the hailstone at time \(t = 0\) is \(u\),

1. find an expression for \(v\) in terms of \(t\). [5]
2. Hence show that the speed of the hailstone approaches the limiting value \(\frac{g}{3k}\). [1]

The variables \(x\) and \(y\) are related by the differential equation $$x^3 \frac{dy}{dx} = xy + x + 1. \qquad (A)$$

1. Use the substitution \(y = u - \frac{1}{x}\), where \(u\) is a function of \(x\), to show that the differential equation may be written as $$x^2 \frac{du}{dx} = u.$$ [4]
2. Hence find the general solution of the differential equation (A), giving your answer in the form \(y = f(x)\). [5]

The differential equation $$\frac{dy}{dx} + \frac{1}{1 - x^2} y = (1 - x)^{\frac{1}{2}}, \quad \text{where } |x| < 1,$$ can be solved by the integrating factor method.

1. Use an appropriate result given in the List of Formulae (MF1) to show that the integrating factor can be written as \(\left(\frac{1 + x}{1 - x}\right)^{\frac{1}{2}}\). [2]
2. Hence find the solution of the differential equation for which \(y = 2\) when \(x = 0\), giving your answer in the form \(y = f(x)\). [6]

1. Find the general solution of the differential equation $$\frac{dy}{dx} - \frac{y}{x} = \sin 2x,$$ expressing \(y\) in terms of \(x\) in your answer. [6]

In a particular case, it is given that \(y = \frac{2}{\pi}\) when \(x = \frac{1}{4}\pi\).

1. Find the solution of the differential equation in this case. [2]
2. Write down a function to which \(y\) approximates when \(x\) is large and positive. [1]

Use the integrating factor method to find the solution of the differential equation $$\frac{\text{d}y}{\text{d}x} + 2y = \text{e}^{-3x}$$ for which \(y = 1\) when \(x = 0\). Express your answer in the form \(y = \text{f}(x)\). [6]

Find the solution of the differential equation $$\frac{dy}{dx} - \frac{x^2y}{1 + x^3} = x^2$$ for which \(y = 1\) when \(x = 0\), expressing your answer in the form \(y = f(x)\). [8]

The variables \(x\) and \(y\) satisfy the differential equation $$\frac{dy}{dx} + 4y = 5 \cos 3x.$$

1. Find the complementary function. [2]
2. Hence, or otherwise, find the general solution. [7]
3. Find the approximate range of values of \(y\) when \(x\) is large and positive. [2]

The substitution \(y = u^k\), where \(k\) is an integer, is to be used to solve the differential equation $$x \frac{dy}{dx} + 3y = x^2 y^2 \qquad (A)$$ by changing it into an equation (B) in the variables \(u\) and \(x\).

1. Show that equation (B) may be written in the form $$\frac{du}{dx} + \frac{3}{kx} u = \frac{1}{k} x u^{k+1}.$$ [4]
2. Write down the value of \(k\) for which the integrating factor method may be used to solve equation (B). [1]
3. Using this value of \(k\), solve equation (B) and hence find the general solution of equation (A), giving your answer in the form \(y = f(x)\). [4]