Methods to solve First Order Differential Equations

What are First Order Differential Equations?

The highest derivative in a first order differential equation is the first derivative (i.e. $\frac{d y}{d x}$ ).

There’s no one size fits all method to solve all kinds of Differential Equations. There are many, and we have to learn through practice which method to use in a given scenario.

But let’s not put the cart before the horse. We should get to know the various methods at our disposal first.

Various Methods for solving First Order Differential Equations

Separation of Variables

In this method, we separate the variables and then integrate.

So, obviously we can use this method only if we can move all x and dx terms to one side of the equation, and all y and dy terms to the other side.

Q. Solve $\frac{d y}{d x} = \frac{k}{y}$ , where k is a constant.

Explanation:

Let’s separate the x and y variables.

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

or y dy = k dx

Now, we can integrate both sides of the above equation separately.

∫ y dy = $\frac{y^{2}}{2}$ + c
(where c is a constant)

∫ k dx = kx + d
(where c is a constant)

So, $\frac{y^{2}}{2}$ + c = kx + d
or $\frac{y^{2}}{2}$ = kx + m
(where m = d - c)

or y = $\sqrt{2 (k x + m)}$

Typical method for solving First Order Linear Differential Equations

What are Linear Differential Equations?

Differential Equation is linear if no variable or derivative in it has exponent other than one (i.e. no $y^{2}$ , no $\sqrt{y}$ , no $\frac{d^{2} y}{d x^{2}}$ , etc). Also, there should be no functions in it, e.g. cos y, log y, etc.

A linear Differential Equation can only have y, and $\frac{d y}{d x}$ .

General form of Linear Differential Equation: $\frac{d y}{d x}$ + F(x) y = G(x)
For example, 2y + 5 $\frac{d y}{d x}$ = 3x

First order differential equation is one in which the highest derivative is the first derivative (i.e.

\frac{d y}{d x}

). It doesn’t matter what exponents/powers y or

\frac{d y}{d x}

have.

To solve First Order Linear Differential Equations, we can use a special method. Let’s see.

We create two new functions of x, called u and v, such that y = uv. Now, we just need to find the values of u and v.

Let’s see this method in detail.

Step 1: As y = uv.
So, $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ (from Product Rule of derivatives)

We will substitute this in the standard form of Linear Differential Equation, i.e. $\frac{d y}{d x}$ + F(x) y = G(x)

Step 2: We separate the parts containing v.

Step 3: Put v term equal to 0.

Step 4: Solve the differential equation in u and x (that we got in the last step), using separation of variables method. Find the value of u.

Step 5: Place the value of u in the equation that we got in Step 2. Solve it to find v.

Step 6: Now, as we know both u and v, just find the value of y (as y = uv).

Things will become clearer if we see an example.

Q. Solve $\frac{d y}{d x} = \frac{y}{x} + 3$

Explanation:

We need to reconstruct the given equation in this form: $\frac{d y}{d x}$ + F(x) y = G(x)

$\frac{d y}{d x} = \frac{y}{x} + 3$
or $\frac{d y}{d x} - \frac{y}{x} = 3$

\frac{d y}{d x} - \frac{y}{x}

means that at many point of the curve, slope -

\frac{y}{x}

will be equal to 3.

Step 1: Substitute y = uv, and $\frac{d y}{d x} = u \frac{d v}{d x} + v \frac{d u}{d x}$ in this equation.

$\frac{d y}{d x} - \frac{y}{x} = 3$
or $u \frac{d v}{d x} + v \frac{d u}{d x} - \frac{u v}{x} = 3$

Step 2: We separate the parts containing v.

$u \frac{d v}{d x} + v \frac{d u}{d x} - \frac{u v}{x} = 3$
or $u \frac{d v}{d x} + v (\frac{d u}{d x} - \frac{u}{x}) = 3$

Step 3: Put v term equal to 0.

So, $\frac{d u}{d x} - \frac{u}{x}$ = 0
or $\frac{d u}{d x} = \frac{u}{x}$

Step 4: Solve the differential equation in u and x (that we got in the last step), using separation of variables method. Find the value of u.

$\frac{d u}{d x} = \frac{u}{x}$
or $\frac{d u}{u} = \frac{d x}{x}$

Now, on integrating them, we get:
ln u = ln x + c (where c is a constant)
or ln u = ln x + ln k (where ln k is a constant)
or u = kx

Step 5: Place the value of u in the equation that we got in Step 2. Solve it to find v.

$u \frac{d v}{d x} + v (\frac{d u}{d x} - \frac{u}{x}) = 3$
or $k x \frac{d v}{d x} = 3$ (v term is already zero, so we can ignore that part.)

Now, solve it using separation of variables method.
$k x \frac{d v}{d x} = 3$
or $k d v = 3 \frac{d x}{x}$

Now, on integrating them, we get:
kv = 3 ln x + ln c
or kv = ln (c $x^{3}$ )
or v = $\frac{l n (c x^{3})}{k}$

Step 6: Now, as we know both u and v, just find the value of y (as y = uv).

y = uv = (kx) $\frac{l n (c x^{3})}{k}$ = x $l n (c x^{3})$

Homogeneous Equations

$\frac{d y}{d x}$ = F ( $\frac{y}{x}$ )

In such a case, we can replace $\frac{y}{x}$ by another variable (say v), and then solve it using Separation of Variables method.

So, v = $\frac{y}{x}$ (i.e. y = vx)

And $\frac{d y}{d x} = \frac{d (v x)}{d x} = v \frac{d x}{d x} + x \frac{d v}{d x} = v + x \frac{d v}{d x}$

Substitute these values in the original equation and solve for v. Now, put v = $\frac{y}{x}$ , and simplify.

Convert Non-linear differential equation into Linear differential equation

We can sometimes convert a non-linear Bernoulli differential equation into a linear differential equation. Then it becomes easy to solve it.

General form of Bernoulli Differential Equation: $\frac{d y}{d x} + F (x) y = G (x) y^{n}$
(Where n can be any real number, except 0 or 1)

We can substitute u = $y^{1 - n}$ in the above equation, and thus turn it into a linear differential equation, which is easier to solve.

$\frac{d y}{d x} + F (x) y = G (x) y^{n}$

When n = 0, then the above equation becomes a Linear Differential Equation.
When n = 1, the the above equation can easily be solved using Separation of Variables method.

In other cases, we can use the method given above.

Q. Solve $\frac{d y}{d x} + x^{3} y = x^{3} y^{3}$

Explanation:

It’s a Bernoulli Differential Equation of the form: $\frac{d y}{d x} + F (x) y = G (x) y^{n}$

Here, F(x) = $x^{3}$ , G(x) = $x^{3}$ , n = 3

So, u = $y^{1 - n} = y^{- 2}$
or y = $u^{- 1 / 2}$

So, $\frac{d y}{d x} = (- 1 / 2) u^{- 3 / 2} \frac{d u}{d x}$

Now, substitute the values of y and $\frac{d y}{d x}$ in the original equation.
$\frac{d y}{d x} + x^{3} y = x^{3} y^{3}$
or $(- 1 / 2) u^{- 3 / 2} \frac{d u}{d x} + x^{3} u^{- 1 / 2} = x^{3} (u^{- 1 / 2})^{3}$

To simplify it, let us multiply the equation by -2 $u^{3 / 2}$ .

$\frac{d u}{d x} - 2 u x^{3} = - 2 x^{3}$

Now, we can easily solve this equation.

$\frac{d u}{d x} = x^{3} (2 u - 2)$
or $\frac{d u}{u - 1} = 2 x^{3} d x$

Now, we can integrate both sides.
$\int \frac{1}{u - 1} d u = \int 2 x^{3} d x$
or ln (u - 1) = 2 $\frac{x^{4}}{4}$ + c (where c is a constant)
or ln (u - 1) = $\frac{x^{4}}{2}$ + c
or u - 1 = $e^{(\frac{x^{4}}{2} + c)}$
or u = $e^{(\frac{x^{4}}{2} + c)}$ + 1

As we know that y = $u^{- 1 / 2}$ , so we can substitute it in the above equation now.

Short Cut

We know that, Bernoulli Differential Equation: $\frac{d y}{d x} + F (x) y = G (x) y^{n}$ , where n ≠ 0 or 1

This can be rewritten as: $\frac{d u}{d x} + (1 - n) u F (x) = (1 - n) G (x)$

Solve it, and then just substitute y = $u^{(\frac{- 1}{n - 1})}$

Exact Equations and Integrating Factors

This method can be used for a first-order differential equation.

In order for us to use this method, we need to find a function I(x, y) whose partial derivatives can be put in place of M and N in the following equation:

M(x, y)dx + N(x, y)dy = 0

That is, $\frac{\partial I}{\partial x} d x + \frac{\partial I}{\partial y}$ dy = 0

Obviously, for this method to work, such a function I(x, y) should exist first of all.

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Last updated on Sep 27, 2021

Methods to solve First Order Differential Equations

What are First Order Differential Equations?

Various Methods for solving First Order Differential Equations

Separation of Variables

Q. Solve dydx=ky, where k is a constant.

Typical method for solving First Order Linear Differential Equations

What are Linear Differential Equations?

Q. Solve dydx=yx+3

Homogeneous Equations

Convert Non-linear differential equation into Linear differential equation

Q. Solve dydx+x3y=x3y3

Exact Equations and Integrating Factors

Q. Solve $\frac{d y}{d x} = \frac{k}{y}$ , where k is a constant.

Q. Solve $\frac{d y}{d x} = \frac{y}{x} + 3$

Q. Solve $\frac{d y}{d x} + x^{3} y = x^{3} y^{3}$