https://file.aiquickdraw.com/mathbot/file/1726232509235hobhe86z.jpg

To solve this problem, we need to find a function \(v(x)\) such that the substitution \(T(x, t) = v(x) + \phi(x, t)\) transforms the given inhomogeneous partial differential equation (PDE) into a homogeneous one in terms of \(\phi(x, t)\).

### Given Inhomogeneous PDE

\[
\frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \frac{\partial T}{\partial x} \right) = 4 + \frac{\partial T}{\partial t} + \frac{\partial^2 T}{\partial t^2}.
\]

### Substitution

Given the substitution \(T(x, t) = v(x) + \phi(x, t)\), we compute the derivatives with respect to \(x\) and \(t\):

1. \(\frac{\partial T}{\partial x} = \frac{dv}{dx} + \frac{\partial \phi}{\partial x}\)
2. \(\frac{\partial^2 T}{\partial x^2} = \frac{d^2v}{dx^2} + \frac{\partial^2 \phi}{\partial x^2}\)
3. \(\frac{\partial T}{\partial t} = \frac{\partial \phi}{\partial t}\)
4. \(\frac{\partial^2 T}{\partial t^2} = \frac{\partial^2 \phi}{\partial t^2}\)

### Substitution into the PDE

Substituting these into the given PDE:

\[
\frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \left( \frac{dv}{dx} + \frac{\partial \phi}{\partial x} \right) \right) = 4 + \frac{\partial \phi}{\partial t} + \frac{\partial^2 \phi}{\partial t^2}.
\]

Expanding the left-hand side:

\[
\frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \frac{dv}{dx} \right) + \frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \frac{\partial \phi}{\partial x} \right) = 4 + \frac{\partial \phi}{\partial t} + \frac{\partial^2 \phi}{\partial t^2}.
\]

To make the PDE homogeneous in \(\phi(x, t)\), the terms that do not involve \(\phi(x, t)\) should cancel out. Therefore, we need:

\[
\frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \frac{dv}{dx} \right) = 4.
\]

### Solving for \(v(x)\)

Let's solve the above ordinary differential equation (ODE) for \(v(x)\):

\[
\frac{\partial}{\partial x} \left( \frac{1}{(3 + 2x^2)} \frac{dv}{dx} \right) = 4.
\]

Integrate with respect to \(x\):

\[
\frac{1}{(3 + 2x^2)} \frac{dv}{dx} = 4x + C_1.
\]

Multiply both sides by \((3 + 2x^2)\):

\[
\frac{dv}{dx} = (4x + C_1)(3 + 2x^2).
\]

Expand and integrate again:

\[
v(x) = \int \left(12x + 8x^3 + 3C_1 + 2C_1x^2 \right) \, dx.
\]

\[
v(x) = 6x^2 + 2x^4 + 3C_1x + C_1x^3 + C_2.
\]

To match the given options, we can set \(C_1 = 0\) and \(C_2 = 0\), resulting in:

\[
v(x) = 6x^2 + 2x^4.
\]

### Conclusion

The correct choice for \(v(x)\) is:

\[
\boxed{v(x) = 6x^2 + 2x^4.}
\]

This corresponds to the answer circled in the image.

Would you like further details or have any questions?

### Related Questions:

1. How do we solve a second-order inhomogeneous differential equation?
2. What is the physical meaning of making a PDE homogeneous?
3. How does the method of separation of variables work in solving PDEs?
4. What are boundary conditions and why are they important in solving PDEs?
5. How do different types of PDEs (elliptic, parabolic, hyperbolic) differ in their solution methods?

**Tip:** Understanding the distinction between homogeneous and inhomogeneous equations is key in solving differential equations, as it often simplifies the problem significantly.

Consider the inhomogeneous partial differential equation in T(x,t) given by: (d/dx) [ (1 / (3 + 2x^2)) (dT/dx) ] = 4 + (dT/dt) + (d²T/dt²). Choose v(x) such a way that the substitution T(x,t) = v(x) + φ(x,t) gives rise to a homogeneous partial differential equation in φ(x,t).

Learn how to solve an inhomogeneous partial differential equation using substitution to find v(x) that transforms the PDE into a homogeneous form. The solution covers the derivation of v(x) = 6x² + 2x⁴, using advanced calculus techniques, and is suitable for university-level students.

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