Unit 1

Partial differential equation

Question-1: Form a partial differential equation from-

Sol.

Here we have-

It contains two arbitrary constants a and c

Differentiate the equation with respect to p, we get-

Or

Now differentiate the equation with respect to q, we get-

Now eliminate ‘c’,

We get

Now put z-c in (1), we get-

Or

Question-2: Solve the differential equation-

Sol: Given the boundary condition that-

At x = 0,

Sol.

Here we have-

On integrating partially with respect to x, we get-

Here f(y) is an arbitrary constant.

Now form the boundary condition-

When x = 0,

Hence-

On integrating partially w.r.t.x, we get-

Question-3: 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

Question-4: Solve-

Sol.

We have-

Then the auxiliary equations are-

Consider the first two equations only-

On integrating

…….. (2)

Now consider the last two equations-

On integrating we get-

…………… (3)

From equation (2) and (3)-

Question-5: Solve-

Sol.

Let-

That means-

Put these values of p and q in

Question-6: Solve-

Sol.

Let
 

Charpit’s subsidiary equations are-

So that- dq = 0 or q = a

 On putting q = a in (1) we get-

Such that-

Integrating

Or

Which is the required solution.

Question-7: Find the characteristics of

Sol.

Here we have-

Now comparing (1) with-

Rr + Ss + Tt + f(x, y, z, p, q) = 0

We get-

R =

Now we see that-

Hence the equation is parabolic.

The is given by-

Putting the values of R, S, and T in this, we get-

Now simplifying (2), we get-

Solving it we get the repeated roots given by

Therefore we get only ne family of the characteristic of (1).

The characteristic equation of (1) is-

On integrating, we get-

log y – log x = log

Which is the required family of characteristics and it represents a family of straight lines passing through the origin.

Question-8: Solve

Solution.

Its auxiliary equation is

The required solution is

Question-9: Solve

Solution.

Given equation in symbolic form is

(D3-3D2 D'+4D'3)z=ex+2y

It’s A.E. is where m = -1,2, 2

Put, D=1,D’=2 

Hence the complete solution is

Question-10: Solve

Sol.

A.E. is

C.F. =

It is a case of failure.

Now,

Question-11: Solve

Solution.

A.E. is

 Hence the complete solution is

Question-12: Solve

Solution.

Hence the complete solution is

Question-13: Solve-

Sol.

The above PDE can be written as-

Now put

Which gives-

Then the equation (1) can be written as-

Hence the required solution will be-