First-order differential equations are equations involving some unknown function and its first derivative. The main purpose of this Calculus III review article is to discuss the properties of solutions of first-order differential equations and to describe some effective methods for finding solutions.

Standard Form

The standard form for a first-order differential equation for the unknown function y(t) is:

\dfrac{dy}{dt} = f(t,y)

Here, f is some function of two variables. Many, but not all, first-order differential equations can be written in standard form by algebraically solving for \dfrac{dy}{dt} and then setting f(t,y) equal to the right side of the resulting equation.

Any differentiable function y = y(t) that satisfies this equation for all t in some interval is called a solution. Some differential equations have no solutions, whereas other differential equations have infinitely many solutions. It is also possible that a differential equation has exactly one solution. The general solution of a differential equation is the set of all solutions. A differential equation along with a subsidiary condition y(t_0) = y_0 , given at some value of the independent variable t = t_0 , constitutes an initial value problem. A solution to an initial value problem is a function y(t) that both solves the differential equation and satisfies the given subsidiary condition y(t_0) = y_0 .

Equations with Separable Variables

There is no universally applicable procedure for solving first-order differential equations in standard form with an arbitrary f(t,y) . Here, we consider a subset of first-order equations that can be directly integrated.

This is possible if the function

f(t,y)

can be presented in the form

f(t,y) = g(t) h(y)

Here, g is a function of only t , and h is a function of only y . The differential equation

\dfrac{dy}{dt} = g(t) h(y)

is said to be separable. We can write it in the differential form

\dfrac{dy}{h(y)} - g(t) dt = 0

The general solution of this equation is given by the following integral:

\int \dfrac{dy}{h(y)} - \int g(t) dt = C

Here, C represents the arbitrary constant of integration. The integrals obtained in this expression may be impossible to evaluate. In such a case, numerical methods can be used to obtain an approximate solution. Even if the indicated integrations can be performed, it may not be algebraically possible to solve for y explicitly in terms of t . In that case, the solution is left in implicit form.

Example 1

Let’s solve the equation:

\dfrac{dy}{dt} = \dfrac{t^2+2}{y}

This first-order differential equation is separable, with

g(t) = t^2+2 and h(y) = \dfrac{1}{y} .

Its solution is

\int y dy - \int (t^2+2) dt = C

Therefore:

\dfrac{y^2}{2} - \dfrac{t^3}{3} - 2 t = C

Solving for the function  y  explicitly, we obtain two solutions:

y = \sqrt{\dfrac{2t^3}{3} + 4t + K} \qquad \text{and} \qquad y = -, \sqrt{\dfrac{2t^3}{3} + 4t + K}

The constant  K  is related to the constant

C via K=2C.

Example 2

Consider the following initial value problem:

\dfrac{dy}{dt} = \dfrac{y cos t}{1+2y^2} ,, \qquad y(0) = 1

To find the solutions, assume that y neq 0 and write the differential equation in the form:

\int \dfrac{1+2y^2}{y} ,dy - \int cos t, dt = 0

Then, integrating the left side with respect to y and the right side with respect to t , we obtain:

ln|y| + y^2 = sin t + C

To satisfy the initial condition, we substitute t=0 and y=1 ; this gives C=1 . Hence, we can present a solution in the form:

ln|y| + y^2 = sin t + 1

Homogeneous Equations

A differential equation in standard form

\dfrac{dy}{dt} = f(t,y)

is homogeneous if

f(\alpha t,\alpha y) = f(t,y)

for any real number \alpha .

Such an equation can always be transformed into a separable equation by the change of an independent variable

y = t,z

along with its corresponding derivative:

\dfrac{dy}{dt} = t ,\dfrac{dz}{dt} + z

The resulting equation with the variables z and t

can be solved as a separable differential equation, because the function f

after such a substitution appears to be a function with a single variable, z .

Let’s illustrate this method through examples.

Example 1

Consider the equation:

\dfrac{dy}{dt} = \dfrac{y+t}{t}

First, we verify the condition for homogeneity:

f(\alpha t,\alpha y) = \dfrac{\alpha y+\alpha t}{\alpha t} = \dfrac{y+t}{t} = f(t, y)

Second, we introduce a new dependent variable,

z , so that z=y/t :

\dfrac{dy}{dt} = t ,\dfrac{dz}{dt} + z = z+1

The equation

t,\dfrac{dz}{dt} = 1

is a first-order differential equation with separable variables, which can be directly integrated:

z = ln|Kt|

Here, we have set the constant of integration

C=-,ln|K| ,

and have noted that

ln|t| + ln|K| = ln|Kt| .

Finally, substituting

z=y/t

back into this solution, we obtain the solution to the original differential equation as

y=tln|Kt| .

Example 2

The following equation is also homogeneous:

\dfrac{dy}{dt} = \dfrac{t^2+3y^2}{2ty}

For the variable

z=y/t ,

we have:

t ,\dfrac{dz}{dt} = \dfrac{1+z^2}{2z}

Integrating, we get the following solution:

z^2 + 1 = C t,,\qquad C = \text{constant}

Now, replacing  z by y/t ,

we obtain the solution for the original equation:

y = C t^3 - t^2

Linear Equations with Variable Coefficients

Consider a first-order differential equation in standard form:

\dfrac{dy}{dt} = f(t,y) .

A differential equation is linear if  f(t,y)

can be written as a function of  t times y,

plus another function of

t : f(t,y) = - ,p(t) y + q(t) .

Consequently, a linear differential equation can always be expressed as:

\dfrac{dy}{dt} + p(t) y = q(t)

Here,   p and q

are given functions of the independent variable t .

First-order linear differential equations cannot be solved by straightforward integration methods,because the variables are not separable.As a result, we need to use a different method of solution. The first step is to multiply the linear differential equation by an undetermined function, \mu(t) :

\mu(t) ,\dfrac{dy}{dt} + \mu(t) p(t) y = \mu(t) q(t)

The question now is whether we can choose \mu(t) so that the left side of this equation is recognizable as the derivative of some particular expression. Let’s note the following equalities:

\dfrac{d}{dt} [\mu(t)y] = \mu(t) \dfrac{dy}{dt} + \dfrac{d\mu(t)}{dt},y = \mu(t) ,\dfrac{dy}{dt} + \mu(t) p(t) y

Here, the second equality is valid under the condition that \mu(t) satisfies the equation:

\dfrac{d\mu(t)}{dt} = p(t) \mu(t)

This is a first-order differential equation with separable variables that can be directly integrated:

\int \dfrac{d\mu}{\mu} - \int p(t) dt = 0

If we assume temporarily that \mu(t) is positive, then we have:

ln\mu(t) = \int p(t) dt + K

By choosing the arbitrary constant K to be zero, we obtain the simplest possible function for \mu . Namely:

\mu(t) = exp\left(\int p(t) dt\right)

Note that the integrating factor \mu(t)

is positive for all  t ,

as we had assumed. Returning to the linear differential equation, we have:

\dfrac{d}{dt} [\mu(t)y] = \mu(t) q(t)

Hence, the general solution is:

y = \dfrac{\int \mu(s) q(s) ds + C}{\mu(t)}

Observe that two integrations are required to find the solution of a linear differential equation: one to obtain the integrating factor \mu(t) , and the other to obtain y .

We will now consider some examples.

Example 1

Let’s solve the differential equation:

\dfrac{dy}{dt} + \dfrac{4}{t},y = t^4

This differential equation is a first-order linear differential equation, with p(t) = 4/t and q(t) = t^4 . We can determine

\int p(t) dt = \int \dfrac{4}{t} ,dt = 4 ln|t| = ln t^4

So, the integrating factor in this case is:

\mu(t) = exp\left(\int p(t) dt\right) = e^{ln t^4} = t^4

Multiplying the differential equation by the integrating factor, we get:

t^4 \dfrac{dy}{dt} + 4 t^3 ,y = t^8 \qquad \text{or} \qquad \dfrac{d}{dt} (t^4 y) = t^8

Integrating both sides of this last equation with respect to t , we obtain the solution:

y = \dfrac{t^5}{9} + \dfrac{C}{t^4}

Example 2

Consider the following initial value problem:

t, \dfrac{dy}{dt} + 2y = 4 t^2,, \qquad y(1) = 2

Dividing both sides by t , we can write this equation in standard form (for linear equations):

\dfrac{dy}{dt} + \dfrac{2}{t},y = 4 t,, \qquad y(1) = 2

Thus, p(t) = 2/t and q(t) = 4 t . To solve this differential equation, we first compute the integrating factor:

\mu(t) = exp\left(\int \dfrac{2}{t} dt\right) = e^{2ln |t|} = t^2

Uponmultiplying the equation by \mu(t) = t^2 , we obtain

t^2, \dfrac{dy}{dt} + 2ty = \dfrac{d}{dt} (t^2 y) = 4 t^3

Therefore:

y = t^2 + \dfrac{C}{t^2} ,,\qquad C=\text{constant}

is the general solution of the initial differential equation. To satisfy the initial condition y(1) = 2 , it is necessary to choose C=1 .

Thus, y = t^2 + \dfrac{1}{t^2}

Bernoulli Equations

In some cases, we can solve a nonlinear equation by changing the dependent variable so that the equation becomes linear. Here, we will investigate the equation that has the form:

\dfrac{dy}{dt} + p(t) y = q(t) y^n

Here, n is supposed to be a real number, restricted by the condition n\neq 0, 1 . Such an equation is called a Bernoulli equation, after Jacob Bernoulli. The substitution z=y^{1-n} reduces a Bernoulli equation to a linear equation. The practical use of this method will be demonstrated in the examples.

Example 1

Let’s solve the differential equation:

\dfrac{dy}{dt} + t y = t y^2

This equation is not linear. It is, however, a Bernoulli differential equation, with p(t)=q(t)=t and n=2 . The required substitution in this case is:

z = y^{1-2} = \dfrac{1}{y},, \qquad \dfrac{dy}{dt} = -,\dfrac{1}{z^2} ,\dfrac{dz}{dt}

Substituting these relations into the initial equation, we obtain:

\dfrac{dz}{dt} - t z = - t

This last equation is linear in the unknown function z . The integrating factor is:

\mu(t) = exp\left(\int(-t)dt \right) = e^{-t^2/2}

Multiplying the differential equation by the integrating factor, we obtain:

\dfrac{d}{dt}\left(z e^{-t^2/2}\right) = -te^{-t^2/2}

The solution of the differential equation is then:

z(t) = C e^{t^2/2} + 1,, \qquad y = \dfrac{1}{z} = \left( C e^{t^2/2} + 1 \right)^{-1}

Example 2

Consider the equation:

\dfrac{dy}{dt} + \dfrac{2y}{t} = \dfrac{y^3}{t^2}

This equation is a Bernoulli differential equation, with

p=2/t , q=1/t^2 and n=3 .

So, we make the substitution:

z = y^{1-3} = \dfrac{1}{y^2},, \qquad \dfrac{dy}{dt} = -,\dfrac{1}{2z^{3/2}} ,\dfrac{dz}{dt}

The differential equation for the variable z is:

\dfrac{dz}{dt} - \dfrac{4}{t} z = - ,\dfrac{2}{t^2}

For this linear differential equation, we can define the following integrating factor:

\mu(t) = exp\left(-,\int\dfrac{4}{t}dt \right) = \dfrac{1}{t^4}

Then, the solution for z(t) is:

\dfrac{d}{dt}\left(\dfrac{z}{t^4}\right) = -,\dfrac{2}{t^6},, \qquad z(t) = \dfrac{2}{5t} + Ct^4,,\qquad C=\text{const}

Finally, we can present the solution for the original differential equation in the form:

y = pm\sqrt{\dfrac{5t}{2+5Ct^5}}

Exact Equations and Integrating Factors

Consider the following first-order differential equation:

m(t,y) + n(t,y) ,\dfrac{dy}{dt} = 0

We suppose it is neither linear nor separable, so the methods suitable for those types of equations are not applicable here. But suppose that we can identify a function Psi(t,y) such that:

\dfrac{\partial Psi}{\partial t} = m(t,y) ,,\qquad \dfrac{\partial Psi}{\partial y} = n(t,y)

Then, the differential equation becomes:

m(t,y) + n(t,y) ,\dfrac{dy}{dt} = \dfrac{\partial Psi}{\partial t} + \dfrac{\partial Psi}{\partial y},\dfrac{dy}{dt} = \dfrac{d}{dt} ,Psi[t, y(t)] = 0

Such a differential equation is called an exact equation. The solution for an exact equation is given implicitly by

Psi(t,y) = C ,

where, as usual, C

represents an arbitrary constant.

A systematic way of determining whether a given differential equation is exact is provided by the following test. If m(t,y) and n(t,y) are continuous functions, then a first-order differential equation of the form:

m(t,y) + n(t,y) ,\dfrac{dy}{dt} = 0

is exact if and only if:

\dfrac{\partial m(t,y)}{\partial y} = \dfrac{\partial n(t,y)}{\partial t}

In some cases, a differential equation that is not exact can be transformed into an exact equation. Such a transformation is possible if we multiply the equation by a suitable integrating factor. To investigate the possibility of implementing this idea more generally, let us multiply the equation by a function \mu ,  then try to choose \mu so that the resulting equation:

\mu(t,y) m(t,y) + \mu(t,y) n(t,y) ,\dfrac{dy}{dt} = 0

passes the test of exactness:

\dfrac{\partial}{\partial y}[\mu m] = \dfrac{\partial}{\partial t}[\mu n]

While in principle, integrating factors are powerful tools for solving differential equations, in practice they can be found only in special cases. The most important situation in which integrating factors can be found occurs when \mu is a function of only one of the variables t or y , instead of both. Assuming that \mu is a function of only t , we have:

\dfrac{d\mu}{dt} = g \mu ,, \qquad g = \dfrac{1}{n}\left( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} \right)

If   g is a function of only t , then the integrating factor \mu can be defined as:

\mu(t) = exp\left(\int g(t) dt\right)

A similar procedure can be used to determine the condition under which the differential equation has an integrating factor dependent only on y . In this case, we find:

\mu(t) = exp\left(-,\int h(t) dt\right),, \qquad \text{where} \qquad \dfrac{1}{m}\left( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} \right) = h(y)

Example 1

Solve the differential equation:

2ty + (1+t^2) \dfrac{dy}{dt} = 0

This equation has the standard form with

m(t,y) = 2ty and n(t,y) = 1+t^2 .

We can easily verify that:

\dfrac{\partial m(t,y)}{\partial y} = \dfrac{\partial n(t,y)}{\partial t} = 2t

So, the differential equation is exact. Thus, there is a Psi(t,y) such that:

\dfrac{\partial Psi}{\partial t} = m (t,y) = 2ty ,, \qquad \dfrac{\partial Psi}{\partial y} = n (t,y) = 1+t^2

Integrating the first of these equations, we obtain:

Psi(t,y) = t^2 y + \varphi(y)

Note that when integrating with respect to t , the constant (with respect to t) of integration can depend on y . We can determine \varphi(y) by differentiating Psi(t,y) with respect to y :

\dfrac{\partial Psi}{\partial y} = t^2 + \dfrac{d\varphi}{dy} = 1+t^2,, \qquad \dfrac{d\varphi}{dy} = 1

Integrating this last equation with respect to

y , we obtain \varphi(y) = y + C_1 ( C_1 = \text{constant} ).

So, we have

Psi(t,y) = (1+t^2) y + C_1

Finally, the solution to the differential equation, which is given implicitly as

Psi(t,y)=C , is:

y = \dfrac{C_2}{1+t^2},,\qquad C_2 = C - C_1

Example 2

Consider the first-order differential equation:

(3ty + y^2) + (t^2+ty) \dfrac{dy}{dt} = 0

This equation has the standard form with

m = 3ty + y^2 and n = t^2+ty .

We can easily calculate:

\dfrac{\partial m(t,y)}{\partial y} - \dfrac{\partial n(t,y)}{\partial t} = (3t+2y) - (2t+y) = t+y

The equation is not exact, but:

g = \dfrac{1}{n}( \dfrac{\partial m}{\partial y} - \dfrac{\partial n}{\partial t} ) = \dfrac{t+y}{t^2+ty} = \dfrac{1}{t}

is a function of t only. Thus, there is an integrating factor \mu that also depends only on t :

\mu(t) = exp\left(\int g(t) dt\right) = exp\left({\int \dfrac{dt}{t}}\right) = t

Multiplying the differential equation by this integrating factor, we obtain:

(3t^2y + ty^2) + (t^3+t^2y) \dfrac{dy}{dt} = 0

We can easily verify that this differential equation is exact. Thus, there is a Psi(t,y) such that:

\dfrac{\partial Psi}{\partial t} = 3t^2y + ty^2 ,, \qquad \dfrac{\partial Psi}{\partial y} = t^3+t^2y

Integrating this equation, we obtain:

Psi(t,y) = t^3 y + \dfrac{1}{2},t^2 y^2 + \varphi(t)

Differentiating Psi(t,y) with respect to t , we have

\dfrac{\partial Psi}{\partial y} = 3t^2y + ty^2 + \dfrac{d\varphi(t)}{dt} = 3t^2y + ty^2 \quad \Rightarrow \quad \dfrac{d\varphi}{dt} = 0

Finally, we can write the solution:

t^3 y + \dfrac{1}{2},t^2 y^2 = C

Wrapping Everything Up

In this calculus III review article, we have investigated different types of first order differential equations that can be taken. Now, you will be able to determine the type of differential equation, then apply the correct method for solving it. We hope this post gives you greater confidence in your knowledge of first order differential equations and facilitates your study of Calculus III.

Let’s put everything into practice. Try this Differential Equations practice question:

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