Partial Differential Equations: Verification, Formation, and Classification

Published on September 19, 2025

Calculus

Solve problems involving partial derivatives, including verifying solutions to PDEs and classifying PDEs based on their discriminant.

Question

Answer

Here’s a step-by-step solution to the math problems presented in the image.

Part (A): Prove that the given function is a solution to the partial differential equation.

i) Given: and the PDE:

First, find the partial derivatives of

with respect to

and

Now, substitute these into the PDE:

Since

, the equation holds true.
Therefore,

is a solution to

ii) Given: and the PDE:

First, find the partial derivatives of

with respect to

and

Now, substitute these into the PDE:

This equation is not generally true for all

and

. For example, if

, then

, which is impossible.
Therefore,

is not a solution to

.
(Self-correction: There might be a typo in the problem statement or my interpretation. Assuming the problem asks to verify if it’s a solution, the answer is no based on the calculations. If it was meant to be , then it would be a solution.)
Let’s re-check the problem statement. It clearly says

. So, the conclusion stands.

Part (B): Find the partial differential equation for each of the following functions.

This part asks to eliminate arbitrary constants to form a PDE.

i) Given:

We have two arbitrary constants,

and

. We need to find at least two partial derivatives to eliminate them.

From

, we have

From

, we have

.
Substitute

into

So,

. This is a PDE involving

and

Let’s try to eliminate both

and

using

and

.
We have

.
From

, substitute

Now, differentiate this with respect to

From

, we have

.
Substitute this into the expression for

Now, differentiate this with respect to

From

, we have

.
Substitute

back into

So, the PDE is:

ii) Given:

We have two arbitrary constants,

and

.
Differentiate with respect to

(Equation 1)

Differentiate with respect to

(Equation 2)

From Equation 1,

.
Substitute

into Equation 2:

Since

, we can divide by

(Equation 3)

Now we need to eliminate

. Differentiate Equation 3 with respect to

(Equation 4)

Differentiate Equation 3 with respect to

(Equation 5)

This approach seems complicated. Let’s try a different strategy.
From

, we have

.
From

, we have

.
Substitute

Now we have expressions for

and

. We need to find a PDE that doesn’t contain

.
Let’s differentiate

with respect to

(Equation A)

Let’s differentiate

with respect to

(Equation B)

Since

(assuming sufficient smoothness), we can equate the

terms.
From (A):

From (B):

So,

This is the same expression for

we found earlier. We still need to eliminate

and

.
Let’s try differentiating

with respect to

So,

Now we have two expressions for

Assuming

and

This is a PDE that does not contain

iii) Given:

We have one arbitrary constant,

. We need to find one partial derivative to eliminate it.

From this,

Substitute

back into the original equation:

So, the PDE is:

To classify a second-order linear PDE of the form

, we look at the discriminant

If , the PDE is hyperbolic.
If , the PDE is parabolic.
If , the PDE is elliptic.

i) Given:

Comparing with

Calculate the discriminant:

Since

, the PDE is parabolic.

ii) Given:

Comparing with

Calculate the discriminant:

Since

, the PDE is hyperbolic.

iii) Given:

Comparing with

Calculate the discriminant:

Since

, the PDE is elliptic.

🚀 Try Free Math Help Now!🏠 Learn More Powerful Features

Partial Differential Equations: Verification, Formation, and Classification

Question

Answer

Related math solutions

Webversion dauerhaft freischalten

Related math solutions