Separable variables - standard (polynomial/exponential x-side)

7 The number of micro-organisms in a population at time \(t\) is denoted by \(M\). At any time the variation in \(M\) is assumed to satisfy the differential equation $$\frac { \mathrm { d } M } { \mathrm {~d} t } = k ( \sqrt { } M ) \cos ( 0.02 t )$$ where \(k\) is a constant and \(M\) is taken to be a continuous variable. It is given that when \(t = 0 , M = 100\).

1. Solve the differential equation, obtaining a relation between \(M , k\) and \(t\).
2. Given also that \(M = 196\) when \(t = 50\), find the value of \(k\).
3. Obtain an expression for \(M\) in terms of \(t\) and find the least possible number of micro-organisms.

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

1. Solve the differential equation and obtain an expression for \(y\) in terms of \(x\).
2. Explain briefly why \(x\) can only take values less than 1 .

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

4 The variables \(x\) and \(y\) are related by the differential equation $$\left( x ^ { 2 } + 4 \right) \frac { d y } { d x } = 6 x y$$ It is given that \(y = 32\) when \(x = 0\). Find an expression for \(y\) in terms of \(x\).

10 \includegraphics[max width=\textwidth, alt={}, center]{3621a7e5-a3fb-42c1-828d-7068fddbf2f9-3_677_691_781_724} A particular solution of the differential equation $$3 y ^ { 2 } \frac { \mathrm {~d} y } { \mathrm {~d} x } = 4 \left( y ^ { 3 } + 1 \right) \cos ^ { 2 } x$$ is such that \(y = 2\) when \(x = 0\). The diagram shows a sketch of the graph of this solution for \(0 \leqslant x \leqslant 2 \pi\); the graph has stationary points at \(A\) and \(B\). Find the \(y\)-coordinates of \(A\) and \(B\), giving each coordinate correct to 1 decimal place.

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

1. (a) Prove by differentiation that

$$\frac { \mathrm { d } } { \mathrm {~d} y } ( \ln \tan 2 y ) = \frac { 4 } { \sin 4 y } , \quad 0 < y < \frac { \pi } { 4 }$$ (b) Given that \(y = \frac { \pi } { 6 }\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \cos x \sin 4 y , \quad 0 < y < \frac { \pi } { 4 }$$ Give your answer in the form \(\tan 2 y = A \mathrm { e } ^ { B \sin x }\), where \(A\) and \(B\) are constants to be determined.

2. (a) Find \(\int \frac { 4 x + 3 } { x } \mathrm {~d} x , \quad x > 0\) (b) Given that \(y = 25\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + 3 ) y ^ { \frac { 1 } { 2 } } } { x } \quad x > 0 , y > 0$$ giving your answer in the form \(y = [ \mathrm { g } ( x ) ] ^ { 2 }\).
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5. (a) Find \(\int \frac { 9 x + 6 } { x } \mathrm {~d} x , x > 0\).
(b) Given that \(y = 8\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 9 x + 6 ) y ^ { \frac { 1 } { 3 } } } { x }$$ giving your answer in the form \(y ^ { 2 } = \mathrm { g } ( x )\). \section*{LU}

1. Given that \(y = 2\) at \(x = \frac { \pi } { 8 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y ^ { 2 } } { 2 \sin ^ { 2 } 2 x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-17_81_102_2649_1779}

1. Given that \(y = 2\) at \(x = \frac { \pi } { 4 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 } { y \cos ^ { 2 } x }$$

1. Given that \(y = 2\) when \(x = - \frac { \pi } { 8 }\), solve the differential equation

$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y ^ { 2 } } { 3 \cos ^ { 2 } 2 x } \quad - \frac { 1 } { 2 } < x < \frac { 1 } { 2 }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

2. Find the particular solution of the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 y ^ { 2 } } { \sqrt { 4 x + 5 } } \quad x > - \frac { 5 } { 4 }$$ for which \(y = \frac { 1 } { 3 }\) at \(x = - \frac { 1 } { 4 }\) giving your answer in the form \(y = \mathrm { f } ( x )\) (6)

7 Solve the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { y }\) given that when \(x = 1 , y = 2\).

5. Given that \(y = - 2\) when \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } \sqrt { x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\).

7. A mathematician is selling goods at a car boot sale. She believes that the rate at which she makes sales depends on the length of time since the start of the sale, \(t\) hours, and the total value of sales she has made up to that time, \(\pounds x\). She uses the model $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k ( 5 - t ) } { x }$$ where \(k\) is a constant.
Given that after two hours she has made sales of \(\pounds 96\) in total,

1. solve the differential equation and show that she made \(\pounds 72\) in the first hour of the sale. The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than \(\pounds 10\) per hour.
2. Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left.

3 A curve satisfies the differential equation \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } y\), and passes through the point \(( 1,1 )\). Find \(y\) in terms of \(x\).

1. (a) Find the general solution of the differential equation

$$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } y } { \mathrm {~d} x } + x y - x = 0$$ giving your answer in the form \(y = \mathrm { f } ( x )\).
(b) Find the particular solution for which \(y = 2\) when \(x = 3\)

10

1. Write down the derivative of \(\sqrt { y ^ { 2 } + 1 }\) with respect to \(y\).
2. Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( x - 1 ) \sqrt { y ^ { 2 } + 1 } } { x y }\) and that \(y = \sqrt { \mathrm { e } ^ { 2 } - 2 \mathrm { e } }\) when \(x = \mathrm { e }\),
   find a relationship between \(x\) and \(y\).

7 The gradient of a curve at the point \(( x , y )\), where \(x > - 2\), is given by $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 3 y ^ { 2 } ( x + 2 ) }$$ The points \(( 1,2 )\) and \(( q , 1.5 )\) lie on the curve. Find the value of \(q\), giving your answer correct to 3 significant figures.

4 Solve the differential equation $$\mathrm { e } ^ { 2 y } \frac { \mathrm {~d} y } { \mathrm {~d} x } + \tan x = 0 ,$$ given that \(x = 0\) when \(y = 0\). Give your answer in the form \(y = \mathrm { f } ( x )\).

7 The gradient of the curve \(y = \mathrm { f } ( x )\) is given by the differential equation $$( 2 x - 1 ) ^ { 3 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + 4 y ^ { 2 } = 0$$ and the curve passes through the point \(( 1,1 )\). By solving this differential equation show that $$f ( x ) = \frac { a x ^ { 2 } - a x + 1 } { b x ^ { 2 } - b x + 1 }$$ where \(a\) and \(b\) are integers to be determined.

10. Given that \(y = 8\) at \(x = 1\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 12 x + 9 ) y ^ { \frac { 1 } { 3 } } } { x }$$ Giving your answer in the form \(y ^ { 2 } = \mathrm { f } ( x )\).

1. The height above ground, \(H\) metres, of a passenger on a roller coaster can be modelled by the differential equation

$$\frac { \mathrm { d } H } { \mathrm {~d} t } = \frac { H \cos ( 0.25 t ) } { 40 }$$ where \(t\) is the time, in seconds, from the start of the ride. Given that the passenger is 5 m above the ground at the start of the ride,

1. show that \(H = 5 \mathrm { e } ^ { 0.1 \sin ( 0.25 t ) }\)
2. State the maximum height of the passenger above the ground. The passenger reaches the maximum height, for the second time, \(T\) seconds after the start of the ride.
3. Find the value of \(T\).

7

1. 1. Solve the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\) to find \(x\) in terms of \(t\).
   2. Given that \(x = 1\) when \(t = 0\), show that the solution can be written as $$x = ( a - \cos b t ) ^ { 2 }$$ where \(a\) and \(b\) are constants to be found.
2. The height, \(x\) metres, above the ground of a car in a fairground ride at time \(t\) seconds is modelled by the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\). The car is 1 metre above the ground when \(t = 0\).
   1. Find the greatest height above the ground reached by the car during the ride.
   2. Find the value of \(t\) when the car is first 5 metres above the ground, giving your answer to one decimal place.