UNIT-4

UNIT-4

Partial differential equation: First order

Question-1: Find the differential equation of all circles touching the y-axis at the origin?

The equation of the circle that touches y-axis at the origin is

….(i)

Differentiating (i) with respect to x, we get

Or

Or

Or

Or 

Or

This is the required differential equation.

Question-2: 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

Question-3: Solve-

Sol.

Here we have-

Integrate w.r.t. x, we get-

Integrate w.r.t. x, we get-

Integrate w.r.t. y, we get-

Question-4: Solve the differential equation-

Given that when y = 0, and u = when x = 0.

Sol.

We have-

Integrating partially w.r.t. y, we get-

Now from the boundary conditions,

Then-

From which,

It means,

On integrating partially w.r.t. x givens-

From the boundary conditions, u = when x = 0

From which-

Therefore the solution of the given equation is-

Question-5: 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-6: 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

Question-7: 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

Question-8: 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-

Question-9: 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-10: Solve-

Sol.

Let

Charpit’s subsidiary equations are-

From the first and fourth ratios,

Substituting p = a – x in the given equation, we get-

So that-

Multiply both sides by ,

Integrating-

Or

Which is the required solution.

UNIT-4

Partial differential equation: First order

Question-1: Find the differential equation of all circles touching the y-axis at the origin?

The equation of the circle that touches y-axis at the origin is

….(i)

Differentiating (i) with respect to x, we get

Or

Or

Or

Or 

Or

This is the required differential equation.

Question-2: 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

Question-3: Solve-

Sol.

Here we have-

Integrate w.r.t. x, we get-

Integrate w.r.t. x, we get-

Integrate w.r.t. y, we get-

Question-4: Solve the differential equation-

Given that when y = 0, and u = when x = 0.

Sol.

We have-

Integrating partially w.r.t. y, we get-

Now from the boundary conditions,

Then-

From which,

It means,

On integrating partially w.r.t. x givens-

From the boundary conditions, u = when x = 0

From which-

Therefore the solution of the given equation is-

Question-5: 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-6: 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

Question-7: 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

Question-8: 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-

Question-9: 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-10: Solve-

Sol.

Let

Charpit’s subsidiary equations are-

From the first and fourth ratios,

Substituting p = a – x in the given equation, we get-

So that-

Multiply both sides by ,

Integrating-

Or

Which is the required solution.