Differential Equations

11 methods · 21 questions

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STEP rewards extracting behaviour before (or instead of) an explicit solution. From the right-hand side of $\frac{d y}{d x} = F (x, y)$ alone you can locate stationary points (where $F = 0$ ), classify them, find limits as $x \to \infty$ , and compare two equations by comparing their gradients, then sketch.

Trigger: the question says "sketch", "describe the behaviour as $t \to \infty$ ", or compares two similar ODEs.

Instances:
(i) set $F = 0$ for stationary points and use the sign of $F$ either side to call maxima/minima;
(ii) take $x \to \infty$ in the (often separable) solution to read off a horizontal asymptote or limiting value;
(iii) compare two integrands/gradients via a pointwise inequality (e.g. $e^{- x^{3}} > e^{- x^{2}}$ on $0 < x < 1$ ) to place one solution curve above the other without solving either;
(iv) match the asymptotic form (e.g. $y \approx 3 - 9/ x$ ) to fix the far-field shape of the sketch.

Linked questions (3)

STEP 2 2018 · Q8

Use the substitution $v = y$ to solve the differential equation $\frac{d y}{d t} = α y^{\frac{1}{2}} - β y (y ⩾ 0, t ⩾ 0),$ where $α$ and $β$ are positive constants. Find the non-constant solution $y_{1} (x)$ that satisfies $y_{1} (0) = 0$ .
Solve the differential equation $\frac{d y}{d t} = α y^{\frac{2}{3}} - β y (y ⩾ 0, t ⩾ 0),$ where $α$ and $β$ are positive constants. Find the non-constant solution $y_{2} (x)$ that satisfies $y_{2} (0) = 0$ .
In the case $α = β$ , sketch $y_{1} (x)$ and $y_{2} (x)$ on the same axes, indicating clearly which is $y_{1} (x)$ and which is $y_{2} (x)$ . You should explain how you determined the positions of the curves relative to each other.

STEP 2 2019 · Q6

Note: You may assume that if the functions

y_{1} (x)

and

y_{2} (x)

both satisfy one of the differential equations in this question, then the curves

y = y_{1} (x)

and

y = y_{2} (x)

do not intersect.

(i) Find the solution of the differential equation

\frac{d y}{d x} = y + x + 1

that has the form

y = m x + c

, where

m

and

c

are constants.

Let

y_{3} (x)

be the solution of this differential equation with

y_{3} (0) = k

. Show that any stationary point on the curve

y = y_{3} (x)

lies on the line

y = - x - 1

. Deduce that solution curves with

k < - 2

cannot have any stationary points.

Show further that any stationary point on the solution curve is a local minimum.

Use the substitution

Y = y + x

to solve the differential equation, and sketch, on the same axes, the solutions with

k = 0

k = - 2

and

k = - 3

.

(ii) Find the two solutions of the differential equation

\frac{d y}{d x} = x^{2} + y^{2} - 2 x y - 4 x + 4 y + 3

that have the form

y = m x + c

.

Let

y_{4} (x)

be the solution of this differential equation with

y_{4} (0) = - 2

. (Do not attempt to find this solution.)

Show that any stationary point on the curve

y = y_{4} (x)

lies on one of two lines that you should identify. What can be said about the gradient of the curve at points between these lines?

Sketch the curve

y = y_{4} (x)

. You should include on your sketch the two straight line solutions and the two lines of stationary points.

STEP 2 2021 · Q5

(i) Use the substitution

y = (x - a) u

, where

u

is a function of

x

, to solve the differential equation

(x - a) \frac{d y}{d x} = y - x,

where

a

is a constant.

(ii) The curve

C

with equation

y = f (x)

has the property that, for all values of

t

except

t = 1

, the tangent at the point

(t, f (t))

passes through the point

(1, t)

.

(a) Given that

f (0) = 0

, find

f (x)

for

x < 1

. Sketch

C

for

x < 1

. You should find the co-ordinates of any stationary points and consider the gradient of

C

x \to 1

. You may assume that

z ln ∣ z ∣ \to 0

z \to 0

.

(b) Given that

f (2) = 2

, sketch

C

for

x > 1

, giving the co-ordinates of any stationary points.