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Here you can find several example questions from STEP past papers for you to practice your skills on. There's questions covering formulating your own differential equation, as well as solving first order and second order problems: everything you need to get started. There's even hints to help you along if you need them - just click "show" to reveal them!
Question 7 STEP I 2011
In this question, you may assume that $\ln (1+x) \approx x -\frac12 x^2$
when $\vert x \vert $ is small.
The height of the water in a tank at time $t$ is $h$.
The initial height of the water is $H$ and water flows into the tank at a constant rate.
The cross-sectional area of the tank is constant.
(i) Suppose that water leaks out at a rate proportional to the height of the water in the tank,
and that when the height reaches $\alpha^2 H$, where $\alpha$ is a constant greater than 1,
the height remains constant.
Show that $$ \frac {d h}{d t } = k( \alpha^2 H -h)\,$$
for some positive constant $k$. Deduce that the time $T$ taken for the water to reach height $\alpha H$ is given by
\[
kT = \ln \left(1+\frac1\alpha\right)\,,
\]
and that $kT\approx \alpha^{-1}$ for large values of $\alpha$.
Hint: What form must the differential equation take for the derivative to be stationary at the required height? How do you solve a differential equation of this type? Can you solve for \(T\) at the height \(\alpha H\)?
(ii) Suppose that the rate at which water leaks out of the tank is proportional to $\sqrt h$ (instead of $h$), and that when the height reaches $\alpha^2 H$, where $\alpha$ is a constant greater than 1, the height remains constant.
Show that the time $T'$ taken for the water to reach height $\alpha H$ is
given by
$$cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln
\left(1+\frac1 {\sqrt\alpha} \right)\right)\,$$
for some positive constant $c$, and that $ cT'\approx \sqrt H$ for large values of $\alpha$.
Hint: What does our differential equation change to? What substitution will let you integrate this separable differential equation? Can you follow the same method again to find \(T'\) for the required height?
Question 7 STEP II 2008
(i) By writing $y=u{(1+x^2)\vphantom{\dot A}}^{\frac12}$, where $u$ is a function of $x$, find the solution of the equation $$\frac 1 y \frac{d y} {d x} = xy + \frac x {1+x^2}$$
for which $y=1$ when $x=0$.
Hint: Can you implicitly differentiate \(y\) to substitute in for \(y\) and \(dy/dx\) in the differential equation? What type of first order differential equation are you left with? How do you solve these?
(ii) Find the solution of the equation $$\frac 1 y \frac{d y} {d x} = x^2y + \frac {x^2 } {1+x^3}$$
for which $y=1$ when $x=0$.
Hint: How can you adapt the substitution method from the first part to the differential equation here?
(iii) Give, without proof, a conjecture for the solution of the equation $$\frac 1 y \frac{d y} {d x} = x^{n-1}y + \frac {x^{n-1} } {1+x^n}$$
for which $y=1$ when $x=0$, where $n$ is an integer greater than 1.
Hint: What pattern can you spot from the solutions to the two previous problems? How would this generalise to the case here?
Question 6 STEP I 2010
Show that, if $y=e^x$, then $$(x-1) \frac{d^2 y}{d x^2} -x \frac{d y}{d x} +y=0\,.\tag{*}$$
In order to find other solutions of this differential equation, now let $y=ue^x$, where $u$ is a function of $x$. By substituting this into $(*)$, show that $$(x-1) \frac{d^2 u}{d x^2} + (x-2) \frac{d u}{d x}=0\,.\tag{**}$$
Hint: Can you differentiate \(y\) twice to substitute into \((*)\)? What about for \(y=ue^x\)?
By setting $ \dfrac {d u}{d x}= v$ in $(**)$ and solving the resulting first order differential equation for $v$, find $u$ in terms of $x$.
Hence show that $y=Ax + Be^x$ satisfies $(*)$, where $A$ and $B$ are any constants.
Hint: If you substitute in for \(v\) what type of first order differential equation do you have? Can you solve back then for \(y\)?