Unit – 2

First order partial differential equations

Q1) What are partial differential equations?

A1)

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.

Q2) Solve

A2)

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

Q3) Solve

A3)

We have

Let z = ax + by + c  ... (2)

On substituting the values of p and q in (1), we have

Putting the value of b in (2), we get

Q4) Solve

A4)

Let u = x + by

So that

Substituting these values of p and q in the given equation, we have

Q5) Solve- x2p2 + y2q2 = z2

A5)

This equation can be transformed as-

x2/z2  p2 + y2/z2  q2 = 1

(x/z . z/x)2 + (y/Z . z/y)2 = 1

Let

z/z = Z,   x/x = X,     y/y = Y

Log z = Z,         log x = X,            log y = Y

Equation (1) can be written as-

P2 + Q2 = 1        ……….(2)

Let the required solution be-

Z = aX + bY + c

P = Z/X = a,        Q = Z/Y = b

From (2) we have-

a2 + b2 = 1    or  b =

Z = aX + Y + c

Log z = a log x + log y + c

Q6) Solve-

A6)

Let-

That means-

p = x2 +c             and               q = c – y2

Put these values of p and q in

Dz = pdx + qdy = (x2 + c) dx + (c – y2) dy

z = (x3/x + cx) + (cy – y3/3) + c1

Q7) Solve

A7)

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

Q8) Solve

A8)

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

Q9) Solve

A9)

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

Q10) Solve (

A10)

Here we have

(

The auxiliary equations are

Or

Integrating first members of (2), we have

Log (x – y) = log (y –z) + log c1

Similarly from last two members of (2), we have

The required solution is

Q11) Solve-

A11)

We have-

Then the auxiliary equations are-

Consider first two equations only-

On integrating

x2/2  = - y2/2 + C1/2                         x2 + y2 = C1

…….. (2)

Now consider last two equations-

On integrating we get-

y2 = yz + C2

y2 – yz = C2

…………… (3)

From equation (2) and (3)-

x2 + y2 = f(y2 – yz)

Q12) Using the method of separation of variables, solve

A12)

Let, u = X(x). T (t).                     (2)

Where X is a function of x only and T is a function of t only.

Putting the value of u in (1), we get

(a) 

On integration log X = cx + log a = log

(b)

On integration

Putting the value of X and T in (2) we have

But,

i.e.

Putting the value of a b and c in (3) we have

Which is the required solution.

Q13) Using the method of separation of variables, solve Where

A13)

Assume the given solution

Substituting in the given equation, we have

Solving (i)

From (ii)

Thus

Now,

Substituting these values in (iii) we get

Which is the required solution