1.08k Separable differential equations: dy/dx = f(x)g(y)

3 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { 2 } + \frac { k } { x ^ { 3 } }\) and passes through the point \(P ( 1,9 )\). The gradient of the curve at \(P\) is 2 .

1. Find the value of the constant \(k\).
2. Find the equation of the curve.

7 In a certain chemical process a substance is being formed, and \(t\) minutes after the start of the process there are \(m\) grams of the substance present. In the process the rate of increase of \(m\) is proportional to \(( 50 - m ) ^ { 2 }\). When \(t = 0 , m = 0\) and \(\frac { \mathrm { d } m } { \mathrm {~d} t } = 5\).

1. Show that \(m\) satisfies the differential equation $$\frac { \mathrm { d } m } { \mathrm {~d} t } = 0.002 ( 50 - m ) ^ { 2 }$$
2. Solve the differential equation, and show that the solution can be expressed in the form $$m = 50 - \frac { 500 } { t + 10 }$$
3. Calculate the mass of the substance when \(t = 10\), and find the time taken for the mass to increase from 0 to 45 grams.
4. State what happens to the mass of the substance as \(t\) becomes very large.

7 In a chemical reaction a compound \(X\) is formed from a compound \(Y\). The masses in grams of \(X\) and \(Y\) present at time \(t\) seconds after the start of the reaction are \(x\) and \(y\) respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time, the rate of formation of \(X\) is proportional to the mass of \(Y\) at that time. When \(t = 0 , x = 5\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 1.9\).

1. Show that \(x\) satisfies the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 0.02 ( 100 - x ) .$$
2. Solve this differential equation, obtaining an expression for \(x\) in terms of \(t\).
3. State what happens to the value of \(x\) as \(t\) becomes very large.

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 \(x\).

8

1. Using partial fractions, find $$\int \frac { 1 } { y ( 4 - y ) } \mathrm { d } y$$
2. Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 4 - y ) ,$$ obtaining an expression for \(y\) in terms of \(x\).
3. State what happens to the value of \(y\) if \(x\) becomes very large and positive.

10 A model for the height, \(h\) metres, of a certain type of tree at time \(t\) years after being planted assumes that, while the tree is growing, the rate of increase in height is proportional to \(( 9 - h ) ^ { \frac { 1 } { 3 } }\). It is given that, when \(t = 0 , h = 1\) and \(\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.2\).

1. Show that \(h\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.1 ( 9 - h ) ^ { \frac { 1 } { 3 } } .$$
2. Solve this differential equation, and obtain an expression for \(h\) in terms of \(t\).
3. Find the maximum height of the tree and the time taken to reach this height after planting.
4. Calculate the time taken to reach half the maximum height.

8 \includegraphics[max width=\textwidth, alt={}, center]{20893bfc-3300-4205-9d2c-729cc3243971-3_597_951_1471_598} In the diagram the tangent to a curve at a general point \(P\) with coordinates \(( x , y )\) meets the \(x\)-axis at \(T\). The point \(N\) on the \(x\)-axis is such that \(P N\) is perpendicular to the \(x\)-axis. The curve is such that, for all values of \(x\) in the interval \(0 < x < \frac { 1 } { 2 } \pi\), the area of triangle \(P T N\) is equal to \(\tan x\), where \(x\) is in radians.

1. Using the fact that the gradient of the curve at \(P\) is \(\frac { P N } { T N }\), show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } y ^ { 2 } \cot x .$$
2. Given that \(y = 2\) when \(x = \frac { 1 } { 6 } \pi\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).

8

1. Express \(\frac { 100 } { x ^ { 2 } ( 10 - x ) }\) in partial fractions.
2. Given that \(x = 1\) when \(t = 0\), solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 100 } x ^ { 2 } ( 10 - x )$$ obtaining an expression for \(t\) 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.

10 The number of birds of a certain species in a forested region is recorded over several years. At time \(t\) years, the number of birds is \(N\), where \(N\) is treated as a continuous variable. The variation in the number of birds is modelled by $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N ( 1800 - N ) } { 3600 }$$ It is given that \(N = 300\) when \(t = 0\).

1. Find an expression for \(N\) in terms of \(t\).
2. According to the model, how many birds will there be after a long time?

6 A certain curve is such that its gradient at a point \(( x , y )\) is proportional to \(x y\). At the point \(( 1,2 )\) the gradient is 4 .

1. By setting up and solving a differential equation, show that the equation of the curve is \(y = 2 \mathrm { e } ^ { x ^ { 2 } - 1 }\).
2. State the gradient of the curve at the point \(( - 1,2 )\) and sketch the curve.

7 The variables \(x\) and \(y\) are related by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 6 x \mathrm { e } ^ { 3 x } } { y ^ { 2 } } .$$ It is given that \(y = 2\) when \(x = 0\). Solve the differential equation and hence find the value of \(y\) when \(x = 0.5\), giving your answer correct to 2 decimal places.

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 \(x\).

5 In a certain chemical process a substance \(A\) reacts with another substance \(B\). The masses in grams of \(A\) and \(B\) present at time \(t\) seconds after the start of the process are \(x\) and \(y\) respectively. It is given that \(\frac { \mathrm { d } y } { \mathrm {~d} t } = - 0.6 x y\) and \(x = 5 \mathrm { e } ^ { - 3 t }\). When \(t = 0 , y = 70\).

1. Form a differential equation in \(y\) and \(t\). Solve this differential equation and obtain an expression for \(y\) in terms of \(t\).
2. The percentage of the initial mass of \(B\) remaining at time \(t\) is denoted by \(p\). Find the exact value approached by \(p\) as \(t\) becomes large.

10 Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, \(t\) minutes later, the volume of liquid in the tank is \(V \mathrm {~cm} ^ { 3 }\). The liquid is flowing into the tank at a constant rate of \(80 \mathrm {~cm} ^ { 3 }\) per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to \(k V \mathrm {~cm} ^ { 3 }\) per minute where \(k\) is a positive constant.

1. Write down a differential equation describing this situation and solve it to show that $$V = \frac { 1 } { k } \left( 80 - 80 \mathrm { e } ^ { - k t } \right)$$
2. It is observed that \(V = 500\) when \(t = 15\), so that \(k\) satisfies the equation $$k = \frac { 4 - 4 e ^ { - 15 k } } { 25 }$$ Use an iterative formula, based on this equation, to find the value of \(k\) correct to 2 significant figures. Use an initial value of \(k = 0.1\) and show the result of each iteration to 4 significant figures.
3. Determine how much liquid there is in the tank 20 minutes after the liquid started flowing, and state what happens to the volume of liquid in the tank after a long time.

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 \(\theta\).

7 Given that \(y = 1\) when \(x = 0\), solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 4 x \left( 3 y ^ { 2 } + 10 y + 3 \right)$$ obtaining an expression for \(y\) in terms of \(x\).

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.

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.

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 \(x = \frac { 1 } { 4 } \pi\).