MTH6131 Chap.16 Separable and Linear First Order Differential Equations
Separable and Linear First Order Differential Equations
Where the course has been heading
The opening lecture describes the course as mathematical modelling with differential equations and lists modelling with ordinary differential equations as the last of its topics.
Its first model, continuously compounded interest, is already one: a balance growing at a rate proportional to itself satisfies dA/dt = rA, whose solution is the exponential function found as a limit in the first lecture.
The schedule places ordinary differential equations in the final lecture, and the exercise sets work through three kinds of task: verifying solutions, solving separable equations and solving linear first-order equations.
Verifying and fixing the constant
To verify that a function solves an equation, differentiate it and substitute; no solving is needed.
A general solution carries an arbitrary constant, and a condition such as a value at one point fixes it, turning the general solution into a particular one.
The equation together with the condition is an initial value problem.
Separable equations
When the derivative factors into a function of x times a function of y, move every y to the side with dy and every x to the side with dx, integrate both sides, and combine the constants.
Dividing by a factor that can vanish may discard a constant solution, which should be stated separately.
Linear equations and the integrating factor
A linear first-order equation is first rewritten in the standard form y′ + P(x)y = Q(x). The exercise solutions then compute the integrating factor V = e to the power of the integral of P, and the solution y = (1/V) times the integral of VQ.
The interval is stated so that absolute values inside a logarithm can be dropped, and an initial condition is applied last. The same method models a fund that earns interest while money is withdrawn, where the equation reveals a balance at which the two exactly cancel.
What this chapter covers
- 01
Differential equations as models of change
- 02
Verifying a proposed solution
- 03
General solutions and initial value problems
- 04
Separating the variables
- 05
Constant solutions lost by division
- 06
Standard form for a linear equation
- 07
The integrating factor and the solution formula
- 08
The interest model with withdrawals
Worked example · free
A separable equation with an initial condition
- 1Separate the variables.
- 2Integrate both sides and combine the constants.
- 2Apply the initial condition and verify.
Key terms
- Differential Equation
- An equation relating an unknown function to one or more of its derivatives.
- Initial Value Problem
- A differential equation together with a condition fixing the value of the solution at one point.
- Separable Equation
- A first-order equation in which the derivative factors into a function of x times a function of y, so the variables can be collected on opposite sides.
- Standard Form
- The arrangement y′ + P(x)y = Q(x) of a linear first-order equation, required before computing the integrating factor.
- General Solution
- A family of solutions containing an arbitrary constant, from which particular solutions are chosen by a condition.
Separable and Linear First Order Differential Equations FAQ
How do I tell whether a first-order equation is separable or linear?
Write the derivative on its own. If the right-hand side factors as a function of x times a function of y, the equation is separable. If it can be rearranged as y′ plus a function of x times y equals a function of x, it is linear. Some equations, such as y′ equals k y, are both, and either method works.
Why must a linear equation be put in standard form first?
Because the integrating factor is built from the coefficient P of y when the coefficient of y′ is exactly one. If the equation has another coefficient on y′, dividing through by it changes both P and Q, and an integrating factor computed from the unrearranged equation will be wrong.
How is this topic connected to the interest model at the start of the course?
The continuously compounded model says the balance grows at a rate proportional to itself, which is the differential equation dA/dt = rA. Solving it by separating variables gives A equals A₀ times e to the power rt, the same function the opening lecture obtained as a limit of compounding more and more often.
Exam move
Classify every equation before touching it, writing separable or linear at the top of the working. Then practise the integrating factor on equations whose coefficient of y′ is not one, because rearranging into standard form is the step most often skipped.
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