Separable variables

6 Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 3 } + 1 } { y ^ { 2 } }$$ obtaining an expression for \(y\) in terms of \(x\).

5 Given that \(y = 0\) when \(x = 1\), solve the differential equation $$x y \frac { \mathrm {~d} y } { \mathrm {~d} x } = y ^ { 2 } + 4 ,$$ obtaining an expression for \(y ^ { 2 }\) in terms of \(x\).

7 The variables \(x\) and \(t\) are related by the differential equation $$\mathrm { e } ^ { 2 t } \frac { \mathrm {~d} x } { \mathrm {~d} t } = \cos ^ { 2 } x$$ where \(t \geqslant 0\). When \(t = 0 , x = 0\).

1. Solve the differential equation, obtaining an expression for \(x\) in terms of \(t\).
2. State what happens to the value of \(x\) when \(t\) becomes very large.
3. Explain why \(x\) increases as \(t\) increases.

5 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { 2 x + y }$$ and \(y = 0\) when \(x = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

8 The variables \(x\) and \(t\) satisfy the differential equation $$t \frac { \mathrm {~d} x } { \mathrm {~d} t } = \frac { k - x ^ { 3 } } { 2 x ^ { 2 } }$$ for \(t > 0\), where \(k\) is a constant. When \(t = 1 , x = 1\) and when \(t = 4 , x = 2\).

1. Solve the differential equation, finding the value of \(k\) and obtaining an expression for \(x\) in terms of \(t\).
2. State what happens to the value of \(x\) as \(t\) becomes large.

4 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 y \mathrm { e } ^ { 3 x } } { 2 + \mathrm { e } ^ { 3 x } }$$ Given that \(y = 36\) when \(x = 0\), find an expression for \(y\) in terms of \(x\).

5 The variables \(x\) and \(\theta\) satisfy the differential equation $$2 \cos ^ { 2 } \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = \sqrt { } ( 2 x + 1 )$$ and \(x = 0\) when \(\theta = \frac { 1 } { 4 } \pi\). Solve the differential equation and obtain an expression for \(x\) in terms of \(\theta\).

9 The number of organisms in a population at time \(t\) is denoted by \(x\). Treating \(x\) as a continuous variable, the differential equation satisfied by \(x\) and \(t\) is $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { x \mathrm { e } ^ { - t } } { k + \mathrm { e } ^ { - t } }$$ where \(k\) is a positive constant.

1. Given that \(x = 10\) when \(t = 0\), solve the differential equation, obtaining a relation between \(x , k\) and \(t\).
2. Given also that \(x = 20\) when \(t = 1\), show that \(k = 1 - \frac { 2 } { \mathrm { e } }\).
3. Show that the number of organisms never reaches 48, however large \(t\) becomes.

4 The variables \(x\) and \(y\) satisfy the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = y \left( 1 - 2 x ^ { 2 } \right)$$ and it is given that \(y = 2\) when \(x = 1\). Solve the differential equation and obtain an expression for \(y\) in terms of \(x\) in a form not involving logarithms.

6 The variables \(x\) and \(\theta\) satisfy the differential equation $$( 3 + \cos 2 \theta ) \frac { \mathrm { d } x } { \mathrm {~d} \theta } = x \sin 2 \theta$$ and it is given that \(x = 3\) when \(\theta = \frac { 1 } { 4 } \pi\).

1. Solve the differential equation and obtain an expression for \(x\) in terms of \(\theta\).
2. State the least value taken by \(x\).

5 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - 2 y } \tan ^ { 2 } x$$ for \(0 \leqslant x < \frac { 1 } { 2 } \pi\), and it is given that \(y = 0\) when \(x = 0\). Solve the differential equation and calculate the value of \(y\) when \(x = \frac { 1 } { 4 } \pi\).

5

1. Differentiate \(\frac { 1 } { \sin ^ { 2 } \theta }\) with respect to \(\theta\).
2. The variables \(x\) and \(\theta\) satisfy the differential equation $$x \tan \theta \frac { d x } { d \theta } + \operatorname { cosec } ^ { 2 } \theta = 0$$ for \(0 < \theta < \frac { 1 } { 2 } \pi\) and \(x > 0\). It is given that \(x = 4\) when \(\theta = \frac { 1 } { 6 } \pi\). Solve the differential equation, obtaining an expression for \(x\) in terms of \(\theta\).

7 The variables \(x\) and \(y\) satisfy the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = x \mathrm { e } ^ { x + y }\). It is given that \(y = 0\) when \(x = 0\).

1. Solve the differential equation, obtaining \(y\) in terms of \(x\).
2. Explain why \(x\) can only take values that are less than 1 .

5 The variables \(x\) and \(y\) satisfy the differential equation $$( x + 1 ) y \frac { \mathrm {~d} y } { \mathrm {~d} x } = y ^ { 2 } + 5$$ It is given that \(y = 2\) when \(x = 0\). Solve the differential equation obtaining an expression for \(y ^ { 2 }\) in terms of \(x\).

6 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = k y ^ { 3 } \mathrm { e } ^ { - x }$$ where \(k\) is a constant. It is given that \(y = 1\) when \(x = 0\), and that \(y = \sqrt { } \mathrm { e }\) when \(x = 1\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

9 Compressed air is escaping from a container. The pressure of the air in the container at time \(t\) is \(P\), and the constant atmospheric pressure of the air outside the container is \(A\). The rate of decrease of \(P\) is proportional to the square root of the pressure difference ( \(P - A\) ). Thus the differential equation connecting \(P\) and \(t\) is $$\frac { \mathrm { d } P } { \mathrm {~d} t } = - k \sqrt { } ( P - A )$$ where \(k\) is a positive constant.

1. Find, in any form, the general solution of this differential equation.
2. Given that \(P = 5 A\) when \(t = 0\), and that \(P = 2 A\) when \(t = 2\), show that \(k = \sqrt { } A\).
3. Find the value of \(t\) when \(P = A\).
4. Obtain an expression for \(P\) in terms of \(A\) and \(t\).

7 The number of insects in a population \(t\) days after the start of observations is denoted by \(N\). The variation in the number of insects is modelled by a differential equation of the form $$\frac { \mathrm { d } N } { \mathrm {~d} t } = k N \cos ( 0.02 t )$$ where \(k\) is a constant and \(N\) is taken to be a continuous variable. It is given that \(N = 125\) when \(t = 0\).

1. Solve the differential equation, obtaining a relation between \(N , k\) and \(t\).
2. Given also that \(N = 166\) when \(t = 30\), find the value of \(k\).
3. Obtain an expression for \(N\) in terms of \(t\), and find the least value of \(N\) predicted by this model.

4 The variables \(x\) and \(\theta\) are related by the differential equation $$\sin 2 \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = ( x + 1 ) \cos 2 \theta$$ where \(0 < \theta < \frac { 1 } { 2 } \pi\). When \(\theta = \frac { 1 } { 12 } \pi , x = 0\). Solve the differential equation, obtaining an expression for \(x\) in terms of \(\theta\), and simplifying your answer as far as possible.

4 During an experiment, the number of organisms present at time \(t\) days is denoted by \(N\), where \(N\) is treated as a continuous variable. It is given that $$\frac { \mathrm { d } N } { \mathrm {~d} t } = 1.2 \mathrm { e } ^ { - 0.02 t } N ^ { 0.5 }$$ When \(t = 0\), the number of organisms present is 100 .

1. Find an expression for \(N\) in terms of \(t\).
2. State what happens to the number of organisms present after a long time.

6 The variables \(x\) and \(y\) are related by the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = 1 - y ^ { 2 }$$ When \(x = 2 , y = 0\). Solve the differential equation, obtaining an expression for \(y\) in terms of \(x\).

7 In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is \(R\) million dollars when the rate of tax is \(x\) dollars per litre. The variation of \(R\) with \(x\) is modelled by the differential equation $$\frac { \mathrm { d } R } { \mathrm {~d} x } = R \left( \frac { 1 } { x } - 0.57 \right)$$ where \(R\) and \(x\) are taken to be continuous variables. When \(x = 0.5 , R = 16.8\).

1. Solve the differential equation and obtain an expression for \(R\) in terms of \(x\).
2. This model predicts that \(R\) cannot exceed a certain amount. Find this maximum value of \(R\).

8 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 5 } x y ^ { \frac { 1 } { 2 } } \sin \left( \frac { 1 } { 3 } x \right)$$

1. Find the general solution, giving \(y\) in terms of \(x\).
2. Given that \(y = 100\) when \(x = 0\), find the value of \(y\) when \(x = 25\).

8 The variables \(x\) and \(\theta\) satisfy the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} \theta } = ( x + 2 ) \sin ^ { 2 } 2 \theta$$ and it is given that \(x = 0\) when \(\theta = 0\). Solve the differential equation and calculate the value of \(x\) when \(\theta = \frac { 1 } { 4 } \pi\), giving your answer correct to 3 significant figures.

6 The variables \(x\) and \(y\) satisfy the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 \cos ^ { 2 } y \tan x ,$$ for \(0 \leqslant x < \frac { 1 } { 2 } \pi\), and \(x = 0\) when \(y = \frac { 1 } { 4 } \pi\). Solve this differential equation and find the value of \(x\) when \(y = \frac { 1 } { 3 } \pi\).

5 The coordinates \(( x , y )\) of a general point on a curve satisfy the differential equation $$x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 2 - x ^ { 2 } \right) y$$ The curve passes through the point \(( 1,1 )\). Find the equation of the curve, obtaining an expression for \(y\) in terms of \(x\).