Concepts: first order linear equations
Differential Equations22 min readLesson 4 of 7
4.1 Standard form and integrating factor
A first order linear equation has the form dy by dx plus P y equal Q, where P and Q depend on x only. Multiply the whole equation by the integrating factor e to the integral of P dx. The left side then becomes the derivative of y times the integrating factor, so integrating both sides is immediate. The solution is y times the factor equal to the integral of Q times the factor, plus a constant. This is why the factor is called integrating: it turns the left side into one derivative.
If the equation is given with a coefficient in front of dy by dx, divide first so that the coefficient is one. If the equation is linear in x rather than in y, write dx by dy plus P1 x equal Q1 and repeat the same method with the roles of x and y exchanged.
4.2 Worked linear equations
Example 1: constant coefficients. For dy by dx plus two y equal four, P is two and the factor is e to the two x. The solution is y e to the two x equal to twice e to the two x plus C, so y equal to two plus C e to the minus two x. The constant is fixed by any initial value.
Example 2: a logarithmic factor. For dy by dx plus y over x equal x squared, P is one over x and the factor is x. Then the left side is the derivative of x y, and integrating gives x y equal x to the fourth over four plus C, so y equal x cubed over four plus C over x.
Example 3: a trigonometric factor. For dy by dx plus y cot x equal cosine squared x, the factor is sine x. The left side becomes the derivative of y sine x, and the right side integrates to minus cosine cubed x over three. The initial condition y of pi by two equal zero then fixes the constant.
4.3 Reading the question for the method
A linear equation is recognised by the presence of y and dy by dx each to the first power, with no product of y and dy by dx and no power of either. When the coefficient P is a constant, the factor is an exponential. When P is a reciprocal of x, the factor is a power of x. When P is cotangent or tangent, the factor is a sine or cosine power. Choosing the factor correctly is most of the work, so write it on its own line and simplify it before using the formula.
Applications include the growth and decay of populations, the discharge of a capacitor, mixing problems and geometric problems where a tangent crosses a curve with a stated property. In each case the modelling step produces a linear equation in the unknown function.
4.4 Common mistakes and exam tip
Using the integrating factor without dividing the equation into standard form is the first error, because a coefficient in front of dy by dx changes P and therefore the factor. Forgetting to multiply Q by the factor before integrating is the second. Writing the constant inside the integral instead of adding it after the integration is the third. In an exam, show the factor, then the left side as one derivative, then the integration; that sequence earns the method marks even if the last algebra slips.
4.5 Solving a linear equation step by step
The steps never change. Write the equation as dy by dx plus P y equal Q, with the coefficient of the derivative equal to one. Compute the integral of P and write the integrating factor as an exponential. Multiply every term by the factor and rewrite the left side as the derivative of y times the factor. Integrate both sides and divide by the factor. Apply the initial condition last, after the constant appears.
A quick numerical check is to substitute x equal to zero into the final solution when the initial condition is given elsewhere; if the numbers do not match the given point, the constant or the factor has an error. The check takes one line and it is especially useful in the exam where the integration step is the most error prone.
4.6 Two patterns worth memorising
When P is a constant k, the integrating factor is e to the k x and the solution is a constant plus a multiple of e to the minus k x. When P is one over x, the factor is x and the left side becomes the derivative of x y. These two cases cover most of the exercise questions and all of the application questions, so recognising them saves a full minute per question.
