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Mathematics (maths) - Partial Differential Equations

**PARTIAL DIFFERENTIAL EQUATIONS**

**1. ****Explain
how PDE are formed? **

PDE can be obtained

(i)
By eliminating the arbitrary constants
that occur in the functional relation between the dependent and independent
variables.

(ii) By
eliminating arbitrary functions from a given relation between the dependent and
independent variables.

**2.From the PDE by
eliminating the arbitrary constants a & b from ***z***
**=*ax*** **+*by***
.**

Given *z* =*ax
*+*by*

Diff. p.w.r. to *x *we
get,

**3. From the PDE by
eliminating the arbitrary constants a & b from**

**5.
Obtain PDE from z =f (sin x + cos y) .**

Given *z* =*f*
(sin *x* +cos *y*) …(1)

Integrating
w.r to *x* on both sides

*z
*= -cos* x c*+

But
z is a function of *x* and *y*

\* z *=-cos* x f*+(*y*)

Hence *c* =*f* ( *y*) .

**7. Mention three types of solution of a
p.d.e (or) Define general and complete integrals of a**

**p.d.e.**

(i) A
solution which contains as many arbitrary constants as there are independent
variables is called a complete integral (or) complete solution.

(ii) A
solution obtained by giving particular values to the arbitrary constants in a
complete integral is called a particular integral (or) particular solution.

(iii)A solution of a p.d.e which
contains the maximum possible number of arbitrary functions is called a general
integral (or) general solution.

This
is of the form F(p,q) = 0.

Hence
the complete integral is *z* =*ax* +*by cz*.

There
is no singular integral.

Taking *c*
=*f*
(*a*)when f is arbitrary.

Eliminating ' *a* ' between (2) & (3) we get the general
solution.

**9. Find the complete integral of **

This equation is of the form z =px
+qy f+(p, q) .

By
Clairaut’s type,put

Therefore the complete integral is

**10.
Find the complete integral of q =2 px
. **

Given q
=2 px .

This equation of the form f (x, p, q) =0 .

**11. Find the complete integral of ***pq***
**=*xy***
.**

Given *pq
*=*xy *.

It is of the form
*f* (*x*, *p* ) =*f*(*y*
, *q*) .

Hence *dz* =*pdx* +*qdy*
.

The given differential equation
can be written as,

Where a & b are arbitrary constant.

To Find The Singular integral:

Diff (1)
p.w.r.to a,

Which is the singular solution.

To Get the general integral:

Put *b* =*f*(*a*)
in (1) , we get

Eliminate a between (5) abd (6) to get the general
solution.

**2.Solve y ^{2}p-xyq=x(z-2y)**

Soln:

Given y^{2}p-xyq=x(z-2y)

This equation of the form Pp+Qq=R

Here, P=y^{2 },Q=-xy , R= x(z-2y)

Hence the general solution is f(x^{2}+y^{2}
, yz-y^{2})=0.

**3.Solve:(3z-4y)p+(4x-2z)q=2y-3x**

Soln:

Given: (3z-4y)p+(4x-2z)q=2y-3x

This equation of the form Pp+q=R

Here, P= (3z-4y)
,Q=(4x-2z) , R= 2y-3x

Again use Lagrangian multipliers 2,3,4,

2x+3y+4z=b.

Hence the general solution is,

F(x^{2}+y^{2}+z^{2} ,
2x+3y+4z)=0.

**4.Find the general solution of x(y ^{2}-z^{2})p+y(z^{2}-x^{2})q=z(x^{2}-y^{2})**

**Soln;**

Given; x(y^{2}-z^{2})p+y(z^{2}-x^{2})q=z(x^{2}-y^{2})

This equation of the form Pp+q=R

Here, P= x(y^{2}-z^{2}) ,Q= y(z^{2}-x^{2})
, R= z(x^{2}-y^{2})

Use Lagrangian multipliers x,y,z,

We get the ratio in (1)

logx +logy+logz=log b

Hence the general solution is,

F(x^{2}+y^{2}+z^{2} , logx +logy+logz)=0.

The auxiliary equation is m^{3}-2m^{2}=0

Replace
D by m
and D’ by 1

m^{2}(m-2)=0

m=0,0 and m=2

=4 *IP* (-*i*(cos(*x* +*y*) +*i* sin (*x* +*y*))

=-4 cos (*x* +*y*)

Hence the general solution is

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