Standard +0.3 This is a standard separable differential equations question with scaffolding. Part (a) gives the partial fractions decomposition needed for part (b), making the integration routine. The algebra to reach the given form requires careful manipulation but follows standard techniques. Part (c) is straightforward substitution and logarithm solving. Slightly above average due to the algebraic manipulation required, but well within typical C3/C4 scope.
13. (a) Express \(\frac { 1 } { ( 4 - x ) ( 2 - x ) }\) in partial fractions.
The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation
where \(k\) is a constant.
(b) solve the differential equation and show that the solution can be written as
$$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$
Given that \(k = 0.1\)
(c) find the value of \(t\) when \(x = 1\), giving your answer, in seconds, to 3 significant figures.
The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$
where \(k\) is a constant.
$$\text { Given that when } t = 0 , x = 0$$
(b) solve the differential equation and show that the solution can be written as
13. (a) Express $\frac { 1 } { ( 4 - x ) ( 2 - x ) }$ in partial fractions.
The mass, $x$ grams, of a substance at time $t$ seconds after a chemical reaction starts is modelled by the differential equation\\
where $k$ is a constant.\\
(b) solve the differential equation and show that the solution can be written as
$$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$
Given that $k = 0.1$\\
(c) find the value of $t$ when $x = 1$, giving your answer, in seconds, to 3 significant figures.
The mass, $x$ grams, of a substance at time $t$ seconds after a chemical reaction starts is modelled by the differential equation
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$
where $k$ is a constant.
$$\text { Given that when } t = 0 , x = 0$$
(b) solve the differential equation and show that the solution can be written as
\hfill \mbox{\textit{Edexcel C34 2018 Q13 [11]}}