Unit - 7

Partial differential equations: First Order

A partial differential equation (PDE) is an equation involving one or more partial derivatives of an (unknown) function, call it u, that depends on two or more variables, often time t and one or several variables in space. The order of the highest derivative is called the order of the PDE. Just as was the case for ODEs, second-order PDEs will be the most important ones in applications.

Just as for ordinary differential equations (ODEs) we say that a PDE is linear if it is of the first degree in the unknown function u and its partial derivatives. Otherwise, we call it nonlinear.

The standard methods of solving the differential equations of the following

Types:

(i) Equations solvable by separation of the variables.

(ii) Homogeneous equations.

(iii) Linear equations of the first order.

(iv) Exact differential equations.

The differential equation of first order and first degree is namely:

Example 1. Solve

Solution. We have,

Separating the variables, we get

(sin y + y cos y) dy = {x (2 log x +1} dx

Integrating both the sides we get

Example 2. Solve the differential equation

Solution.

Put,

Solutions of first order linear PDEs

Linear Equations of the First Order

A linear partial differential equation of the first order, commonly known as Lagrange’s Linear equation is of the for4m

Pp + Qq = R (1)

Where, P, Q and R are functions of x, y, z. This equation is called a quast linear equation. When P, Q and R are independent of z it is known as linear equation.

Such asn equation is obtained by eliminating an arbitrary function from

where u,v are are some functions of x, y, z.

Differentiating (2) partially with respect to x and y

Eliminating and , we get

which simplifies to

This is of the same form as (1)

Now suppose u = a and v=b, where a, b are constants, so that

By cross multiplication we have,

The solution of these equations are u = a and v = b

Therefore, is the required solution of (1).

Thus, to solve the equation Pp + Qq =R.

(i) Form the subsidiary equations

(ii) Solve these simultaneous equations

(iii) Write the complete solution as or u=f(v)

Example. Solve

Solution. Rewriting the given equation as

The subsidiary equations are

The first two fractions give

Integrating we get n (i)

Again, the first and third fraction give xdx = zdz

Integrating, we get

Hence from (i) and (ii), the complete solution is

Example. Solve

Solution. Here the subsidiary equations are

Using multipliers x, y and z we get each fraction =

which on integration gives

Again, using multipliers l, m and n we get each fraction

which on integration gives lx +my +nz = b (ii)

Hence from (i) and (ii) the required solution is

Example. Solve

Solution. Here the subsidiary equations are

From the last two fractions, we have

Which on integration gives log y = log z + log a or y/z=a (i)

Using multipliers x, y and z we have

Each fraction

which on integration gives

Hence from (i) and (ii) the required solution is

Lagrange’s linear equation in an equation of the type-

Here P, Q and R are the functions of x, y, z and p = and q =

Working steps to solve-

Step-1: Write down the auxiliary equation-

Step-2: Solve the auxiliary equations-

Suppose the two solutions are- u = and v =

Step-3: Then f (u, v) = 0 or u = ∅(v) is the required solution of

Method of multipliers-

Let the auxiliary equation be

L, m, n may be the constants of x, y, z then we have-

L, m, n are selected in a such a way that-

Thus

On solving this differential equation, if the solution is- u =

Similarly, choose another set of multipliers and if the second solution is v =

So that the required solution is f (u, v) = 0.

Example: Solve-

Sol.

We have-

Then the auxiliary equations are-

Consider first two equations only-

On integrating

…….. (2)

Now consider last two equations-

On integrating we get-

…………… (3)

From equation (2) and (3)-

Example: Find the general solution of-

Sol. The auxiliary simultaneous equations are-

……….. (1)

Using multipliers x, y, z we get-

Each term of (1) is equals to-

Xdx + ydy + zdz=0

On integrating-

………… (2)

Again equation (1) can be written as-

Or

………….. (3)

From (2) and (3), the general solution is-

Non-linear partial differential equations-

Type-1: Equation of the type f (p, q) = 0

Method-

Let the required solution is-

Z = ax + by + c ……... (1)

So that-

On putting these values in f (p, q) = 0

We get-

f (a, b) = 0

So, from this, find the value of b in terms of a and put the value of b in (1). It will be the required solution.

Type-2: Equation of the type-

Z = px + qy + f (p, q)

Its solution will be-

Z = ax + by + f (a, b)

Type-3: Equation of the type f (z, p, q) = 0

Type-4: Equation of the type-

Method-

Let-

Example: Solve-

Sol.

This equation can be transformed as-

………. (1)

Let

Equation (1) can be written as-

………… (2)

Let the required solution be-

From (2) we have-

Example: Solve-

Sol.

Let u = x + by

So that-

Put these values of p and q in the given equation, we get-

Example: Solve-

Sol.

Let-

That means-

Put these values of p and q in

References:

1. G.B. Thomas and R.L. Finney, “Calculus and Analytic Geometry”, Pearson, 2002.

2. T. Veerarajan, “Engineering Mathematics”, McGraw-Hill, New Delhi, 2008.

3. B. V. Ramana, “Higher Engineering Mathematics”, McGraw Hill, New Delhi, 2010.

4. N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.

5. B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.