What is the solution of dy/dx = k times y minus a?

y equals a plus A e to the k x, a very common form worth memorising.

Why does ln of y become A e to the k x?

Exponentiating ln of y equals k x plus C gives y equals plus or minus e to the C times e to the k x, that is A e to the k x.

What is the solution of dy/dx = k times y minus a?

y equals a plus A e to the k x, a very common form worth memorising.

Why does ln of y become A e to the k x?

Exponentiating ln of y equals k x plus C gives y equals plus or minus e to the C times e to the k x, that is A e to the k x.

Year 12 Maths Extension 1 (2027) Further applications of calculus

Solve dy/dx = g(y)

Q: How do you solve dy/dx = g(y)?

Separate and integrate the integral of one over g of y dy equals x plus C, then solve for y.

Q: How do you handle the constants?

Combine them into a single constant A as you solve for y.

Q: Can this give an inverse trig answer?

Yes; for example dy by dx equals one plus y squared gives arctan y equals x plus C.

20 practice questions 1 video lesson Theory + worked examples

Learn to solve differential equations where the derivative depends only on $y$ in NSW Year 12 Mathematics Extension 1 (NESA outcome ME1-12-05). When the rate of change is a function of $y$ alone, the equation is rearranged and integrated with respect to $y$.

You will learn to integrate the reciprocal with respect to $y$ and, where possible, rearrange to make $y$ the subject — solving equations such as the exponential growth and decay models that recur throughout the Extension 1 course.

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Practice questions

Every question with a fully worked solution.

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Theory

When $\frac{d y}{d x} = g (y)$ depends only on $y$ , separate and integrate: $\int \frac{1}{g (y)} d y = x + C$ . This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-05) includes the common form $\frac{d y}{d x} = k (y - a) \Rightarrow y = a + A e^{k x}$ .

When $\frac{d y}{d x} = g (y)$ depends only on $y$ , separate the variables: gather the $y$ 's with $d y$ and the $x$ 's with $d x$ , then integrate.

This gives $\int \frac{1}{g (y)} d y = \int d x = x + C$ , after which you rearrange for $y$ .

A very common case is $\frac{d y}{d x} = k (y - a)$ , whose solution is $y = a + A e^{k x}$ .

NESA link. Part of the Year 12 Further applications of calculus focus area, outcome ME1-12-05 ("applies calculus to solve problems involving polynomials, further rates of change, areas and volumes and differential equations") with MAO-WM-01. Syllabus examples include $\frac{d y}{d x} = 3 y$ and $\frac{d y}{d x} = 20 e^{- 5 y}$ .

\frac{d y}{d x} = k y

gives an exponential family

y = A e^{k x}

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

gives

y = a + A e^{k x}

, curving away from

y = a

\frac{d y}{d x} = g (y) ⟹ \int \frac{1}{g (y)} d y = \int d x = x + C .

integral of 1/g(y) dy = x + C

The common linear case:

\frac{d y}{d x} = k (y - a) ⟹ y = a + A e^{k x} .

dy/dx = k(y - a) gives y = a + A e^(kx)

Combine constants. After integrating $\frac{1}{y}$ , the $\ln | y | = k x + C$ rearranges to $y = A e^{k x}$ with $A = \pm e^{C}$ .

How to solve $\frac{d y}{d x} = g (y)$

Divide by $g (y)$ and treat $\frac{d y}{d x}$ as a ratio: $\frac{1}{g (y)} d y = d x$ .
Integrate both sides.
Solve for $y$ , combining constants into a single $A$ .
Apply any initial condition to fix $A$ .

Example 1 — Exponential

Find the general solution of

\frac{d y}{d x} = 5 y

Solution

$\int \frac{1}{y} d y$	$=$	$\int 5 d x$
$\ln \| y \|$	$=$	$5 x + C \Rightarrow y = A e^{5 x}$

y = A e^(5x)

$y = A e^{5 x}$ .

Example 2 — Linear form

Solve

\frac{d y}{d x} = 2 (y - 3)

Solution

Of the form $k (y - a)$ with $k = 2$ , $a = 3$ .

y

=

3 + A e^{2 x}

y = 3 + A e^(2x)

$y = 3 + A e^{2 x}$ .

Example 3 — A reciprocal

Find the general solution of

\frac{d y}{d x} = y^{2}

Solution

$\int \frac{1}{y^{2}} d y$	$=$	$\int d x$
$- \frac{1}{y}$	$=$	$x + C \Rightarrow y = - \frac{1}{x + C}$

y = -1/(x + C)

$y = - \frac{1}{x + C}$ .

Example 4 — Giving arctan

Solve

\frac{d y}{d x} = 1 + y^{2}

given

y (0) = 1

Solution

$\int \frac{d y}{1 + y^{2}} = \int d x$ , so $\tan^{- 1} y = x + C$ .

\tan^{- 1} 1 = C

\Rightarrow

C = \frac{π}{4}

y = tan(x + pi/4)

$y = \tan (x + \frac{π}{4})$ .

Common pitfalls

Not separating. Keep

d y

with the

y

-terms and

d x

with the

x

-terms before integrating.

The linear shortcut.

\frac{d y}{d x} = k (y - a) \Rightarrow y = a + A e^{k x}

— recognise it to save work.

Constant handling. Combine into one

A

(often

A = \pm e^{C}

) rather than carrying several.

Absolute values.

\ln | y |

becomes

y = A e^{k x}

with

A

allowed to be negative.

Frequently asked questions

How do you solve dy/dx = g(y)?

Separate and integrate: $\int \frac{1}{g (y)} d y = x + C$ , then solve for $y$ .

What is the solution of dy/dx = k(y - a)?

$y = a + A e^{k x}$ , a very common form worth memorising.

Why does ln|y| become Ae^(kx)?

Exponentiating $\ln | y | = k x + C$ gives $y = \pm e^{C} e^{k x} = A e^{k x}$ .

How do you handle the constants?

Combine them into a single constant $A$ as you solve for $y$ .

Can this give an inverse trig answer?

Yes — e.g. $\frac{d y}{d x} = 1 + y^{2}$ gives $\tan^{- 1} y = x + C$ .

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More in Further applications of calculus

Solve dy/dx = g(y)

📖 Theory

How to solve dydx=g(y)

Common pitfalls

Frequently asked questions

Theory

How to solve $\frac{d y}{d x} = g (y)$