Differential Equations

Linear Differential Equations

A general linear differential equation of order n, in the dependent variable y and the independent variable x, is an equation that can be expressed in the form –

\(a_0(x)\frac{d^ny}{dx^n} + a_1(x)\frac{d^{n – 1}y}{dx^{n – 1}} + ….. + a_{n – 1}(x)\frac{dy}{dx} + a_n(x)y = b(x)\)

where a0 is not identically 0. In this article, we will work out a standard method of solving a practically relevant first order linear differential equation and derive a formula for the general solution. So let’s begin!

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Properties of a General Linear Differential Equation

They possess the following properties as follows:

  • the function y and its derivatives occur in the equation up to the first degree only
  • no products of y and/or any of its derivatives are present
  • no transcendental functions – (trigonometric or logarithmic etc) of y or any of its derivatives occur

A linear differential equation of the first order is a differential equation that involves only the function y and its first derivative. Such equations are physically suitable for describing various linear phenomena in biology, economics, population dynamics, and physics.

Linear Differential Equation

Linear First Order Differential Equations

A linear first order equation is one that can be reduced to a general form – $${\frac{dy}{dx} + P(x)y = Q(x)}$$ where P(x) and Q(x) are continuous functions in the domain of validity of the differential equation. If P(x) or Q(x) is equal to 0, the differential equation can be reduced to a variables separable form which can be easily solved. You can check this for yourselves.

Browse more Topics Under Differential Equations

Integrating Factor

To find the solution of the linear first order differential equation as defined above, we must introduce the concept of an integrating factor. An integrating factor is a term, which when multiplied by an expression, converts it to an exact differential i.e. a function which is the derivative of another function. For example – given an expression,

\(y – x\frac{dy}{dx}\)

If we multiply this expression by \(\frac{1}{x^2}\), note that it becomes an exact differential!

\(\frac{1}{x^2}.(y – x\frac{dy}{dx})\)

\( = \frac{d}{dx}(\frac{y}{x})\)

Solving the Linear First Order Differential Equation

For the general first order linear differential equation, we assume that an integrating factor, that is only a function of x, exists. Let us call it η(x). Then this integrating factor, if multiplied by the expression in the differential equation should reduce the expression to an exact differential. Let us see how –

 \({\frac{dy}{dx} + P(x)y = Q(x)}\)

\(η(x)\frac{dy}{dx} + η(x)P(x)y = η(x)Q(x)\)

On inspection,note that if we put η(x)P(x) = η'(x), we can easily get an exact differential on the left hand side by the

Product Rule of Differentiation

\(η(x)\frac{dy}{dx} + η'(x)y = η(x)Q(x)\)

\(\frac{d}{dx}(η(x)y) = η(x)Q(x)\)

Now integrate on both sides with respect to x –

\(\int{\frac{d}{dx}(η(x)y)dx} = \int{η(x)Q(x)dx}\)

\(η(x)y + c =  \int{η(x)Q(x)dx}\)

\(y = \frac{\int{η(x)Q(x)dx} – c}{η(x)}\)

Thus, we have got our general solution in the terms of our integrating function η(x). You might worry that η(x) is unknown to us. But on the contrary, it is very easy to compute in terms of other known quantities. If you remember, we had used the condition η(x)P(x) = η'(x) to get the above solution. This is actually a variables separable equation which can be easily solved for η(x) in the following way –

\(η(x)P(x) = η'(x)\)

\(P(x) = \frac{η'(x)}{η(x)}\)

\(\int{P(x)dx} = \int{\frac{η'(x)}{η(x)}dx}\)

\(\int{P(x)dx} = ln(η(x)) + k\)

\(η(x) = e^{\int{P(x)dx} – k}\)

\(η(x) = Ke^{\int{P(x)dx}} \), where K = e-k

This expresses the integrating function η(x) in terms of an arbitrary constant K, and the known function P(x). Let us now put this back into  our solution of the differential equation.

\(y = \frac{\int{η(x)Q(x)dx} – c}{η(x)}\)

\(y = \frac{K\int{e^{\int{P(x)dx}}Q(x)dx} – c}{Ke^{\int{P(x)dx}}}\)

\(y = \frac{\int{e^{\int{P(x)dx}}Q(x)dx} – \frac{c}{K}}{e^{\int{P(x)dx}}}\)

Let us just denote the arbitrary constant \(\frac{c}{K}\) as C which is simply another arbitrary constant. In a concise form, we can then write the differential equation along with its solution as –

\({\frac{dy}{dx} + P(x)y = Q(x)}\)

has the solution, \(y = \frac{\int{η(x)Q(x)dx} – C}{η(x)}\)

where \(η(x) = e^{\int{P(x)dx}} \).

Thus, if we express the given differential equation in the standard form, we can now easily write down the solution using the above formula. Look at the solved examples below that make the process clearer.

Equations Reducible to Linear Form: Bernoulli’s Equation

A differential equation of the form –

\(y’ + p(x)y = g(x)y^a\)

where a is a Real Number, is known as the Bernoulli’s Equation. If a = 0, or a =1, it is a straightforward Linear Differential Equation to solve. However, for other values of a, the following method reduces the equation to a linear form –

⇒ Introduce a new dependent variable u(x) such that –

\(u(x) = [y(x)]^{1 – a}\)

⇒ The derivative of u can be found out as –

\(u’ = (1 – a)y^{-a}y’\)

⇒ From the differential equation, the value of y’ can be substituted in the expression for u’ –

\(u’ = (1 – a)y^{-a}(gy^a – py)\)

\( = (1 – a)(g – py^{1 – a})\)

\( = (1 – a)(g – pu)\)

⇒ The expression for the derivative of u can be rearranged to get –

\(u’ + (1 – a)pu = (1 – a)g\)

which is now an equation in the linear form with P(x) = (1 – a)pu, and Q(x) = (1 – a)g. This can then be solved for u(x) in a manner similar to the one for solving Linear Differential Equations. In the end, one must remember to back-substitute and get the final solution i.e. y(x).

Solved Examples For You

Question 1: Solve

\(x\frac{dy}{dx} – 2y = x^3cos4x\)

Answer : The first step is to convert the given differential equation to the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). On rearranging, we can obtain the given differential equation as –

\(\frac{dy}{dx} – \frac{2}{x}y = x^2cos4x\)

On comparing with the standard form, we get \(P(x) = -\frac{2}{x}\) and \(Q(x) = x^2cos4x\). We should now compute the integrating factor first, then proceed towards the general solution. From the formula \(η(x) = e^{\int{P(x)dx}} \), we can work out the integrating factor here as –

\(η(x) = e^{\int{(-\frac{2}{x})dx}}\)

\(= e^{-2lnx}\)

\(= e^{ln(x^{-2}} = x^{-2}\)

Now we can use this in the formula that we had derived for the general solution.

\(y = \frac{\int{η(x)Q(x)dx} – C}{η(x)}\)

\(y = \frac{\int{x^{-2}x^2cos4xdx} – C}{x^{-2}}\)

\(y = x^2(\frac{sin4x}{4} – C)\)

Question 2: Solve the Bernoulli’s Equation

\(y’ = Ay – By^2\), where A and B are constants.

Answer : In the standard form, the equation can be rewritten as –

\(y’ – Ay = -By^2\)

Clearly, a = 2 here, and we must go ahead with the substitution \(u = y^{1 – a} = y^{-1}\). On differentiating u, and substituting the value of y’ from the differential equation, we get –

\(u’ = -y^{-2}y’\)

\( = -y^{-2}(Ay – By^2)\)

Or, \( = B – Ay^{-1}\)

\( = B – Au\)

On rearranging, we obtain our linear differential equation in u as –

\(u’ + Au = B\)

The general solution then is –

\(u = \frac{\int{η(x)Bdx} – C}{η(x)}\)

where \(η(x) = e^{\int{Adx}} = e^{Ax}\)

Then, \(u = \frac{B\int{e^{Ax}dx} – C}{e^{Ax}}\)

\( = \frac{B}{A} – Ce^{-Ax}\)

Finally back-substituting for y, we get –

\(y = \frac{1}{u} = \frac{1}{\frac{B}{A} – Ce^{-Ax}}\)

where C is an arbitrary constant, and A and B are known constants. Note directly from the given equation that y(x) = 0 for all x, is also a solution. It is a singular solution since it can’t be obtained from the general solution.

This brings us to the end of this topic of a linear differential equation.

Question 3: What is the linear first-order differential equation?

Answer: It is an equation where P(x) and Q(x) are two continuous functions in the domain of validity of the differential equation. Furthermore, if P(x) or Q(x) is equal to 0, the differential equation can be reduced to a variable separable form which can be easily solved.

Question 4: What is a non-linear differential equation?

Answer: It refers to a differential equation that is not a linear equation in the unknown function and its derivatives (the linearity or non-linearity in the arguments of the function are not considered here).

Question 5: What makes a nonlinear equation?

Answer: In general, a system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. In addition, any question that cannot be written in this form is nonlinear. Besides, the substitution method that we use for linear systems is the same method we will use for nonlinear systems.

Question 6: What makes a PDE linear?

Answer: A Partial Differentiation Equation in U is categorized as linear if all the terms involving and any of its derivatives can be expressed as a linear combination in which the constants of the U-terms are independent of U.

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