Certainly! Let's solve the given differential equation step by step.
The differential equation given is:
[tex]\[ 3x^2 y \, dx + x^3 \, dy = 0. \][/tex]
First, let's rewrite this equation in the form:
[tex]\[ (3x^2 y) \, dx + (x^3) \, dy = 0. \][/tex]
We want to check if this differential equation is exact. A differential equation [tex]\( M(x, y) \, dx + N(x, y) \, dy = 0 \)[/tex] is exact if:
[tex]\[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. \][/tex]
Here, [tex]\( M(x, y) = 3x^2 y \)[/tex] and [tex]\( N(x, y) = x^3 \)[/tex].
1. Compute [tex]\(\frac{\partial M}{\partial y}\)[/tex]:
[tex]\[ \frac{\partial}{\partial y}(3x^2 y) = 3x^2. \][/tex]
2. Compute [tex]\(\frac{\partial N}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial}{\partial x}(x^3) = 3x^3. \][/tex]
Clearly:
[tex]\[ \frac{\partial M}{\partial y} = 3x^2 \quad \text{and} \quad \frac{\partial N}{\partial x} = 3x^2. \][/tex]
Since these partial derivatives are equal, the differential equation is exact.
To find the solution, we seek a potential function [tex]\( \psi(x, y) \)[/tex] such that:
[tex]\[ \frac{\partial \psi}{\partial x} = M(x, y) = 3x^2 y, \][/tex]
[tex]\[ \frac{\partial \psi}{\partial y} = N(x, y) = x^3. \][/tex]
Integrate [tex]\( \frac{\partial \psi}{\partial x} \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \psi(x, y) = \int 3x^2 y \, dx = y \int 3x^2 \, dx = y \left( x^3 \right) + h(y) = x^3 y + h(y), \][/tex]
where [tex]\( h(y) \)[/tex] is a function of [tex]\( y \)[/tex] only.
Next, differentiate [tex]\( \psi(x, y) \)[/tex] with respect to [tex]\( y \)[/tex] to find [tex]\( h(y) \)[/tex]:
[tex]\[ \frac{\partial \psi}{\partial y} = \frac{\partial}{\partial y} (x^3 y + h(y)) = x^3 + h'(y). \][/tex]
We know from the given [tex]\( N(x, y) \)[/tex] that:
[tex]\[ \frac{\partial \psi}{\partial y} = x^3. \][/tex]
Thus:
[tex]\[ x^3 + h'(y) = x^3. \][/tex]
This implies:
[tex]\[ h'(y) = 0. \][/tex]
Therefore:
[tex]\[ h(y) = C, \][/tex]
where [tex]\( C \)[/tex] is a constant.
Thus, the potential function [tex]\( \psi(x, y) \)[/tex] becomes:
[tex]\[ \psi(x, y) = x^3 y + C. \][/tex]
The solution to the differential equation is given by setting the potential function equal to a constant:
[tex]\[ x^3 y = C'. \][/tex]
Here, [tex]\( C' \)[/tex] represents an arbitrary constant, which we can write as just [tex]\( C \)[/tex] for simplicity.
Therefore, the solution to the differential equation [tex]\( 3x^2 y \, dx + x^3 \, dy = 0 \)[/tex] is:
[tex]\[ x^3 y = C. \][/tex]
This corresponds to option [tex]\( \boxed{D} \)[/tex].
