Q19Mathematics I

Question

Solve the differential equation: $cos x \frac{d ^{2} y}{d x ^{2}} + sin x \frac{d y}{d x} - 2 y cos^{3} x = 2 cos^{3} x$ .

Answer 1

This advanced calculus problem requires us to repeatedly apply a specific mathematical differential operator to a complex multivariable function. The function is defined strictly as:

u (x, y, z) = lo g (x^{3} + y^{3} + z^{3} - 3 x y z)

The differential operator $D$ is formally defined as the exact sum of the first-order partial derivatives with respect to all three spatial variables: $D = \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}$ . We must sequentially prove the exact outcomes of evaluating $D (u)$ and $D^{2} (u)$ .

Part 1: Evaluating the First-Order Operator $D (u)$

Evaluating $D (u)$ physically means we must calculate all three independent partial derivatives of $u$ and sum them up perfectly:

D (u) = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z}

To calculate the partial derivative strictly with respect to $x$ , we employ the standard logarithmic chain rule: $\frac{d}{d x} lo g (f (x)) = \frac{f ^{'} ( x )}{f ( x )}$ .

\frac{\partial u}{\partial x} = \frac{1}{x ^{3} + y ^{3} + z ^{3} - 3 x y z} \cdot \frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} - 3 x y z) = \frac{3 x ^{2} - 3 y z}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}

Because the original argument of the logarithm is completely symmetric with respect to any permutation of $x$ , $y$ , and $z$ , we can immediately and confidently state the other two partial derivatives by exploiting this absolute mathematical symmetry:

\frac{\partial u}{\partial y} = \frac{3 y ^{2} - 3 x z}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}

\frac{\partial u}{\partial z} = \frac{3 z ^{2} - 3 x y}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}

We rigorously sum these three fractional components. Since they conveniently share the exact same denominator, we simply sum their numerators:

D (u) = \frac{( 3 x ^{2} - 3 y z ) + ( 3 y ^{2} - 3 x z ) + ( 3 z ^{2} - 3 x y )}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}

We extract the common numerical factor of $3$ from the entire numerator:

D (u) = 3 \cdot \frac{x ^{2} + y ^{2} + z ^{2} - x y - y z - z x}{x ^{3} + y ^{3} + z ^{3} - 3 x y z}

To simplify this fraction, we must utilize a famous and fundamental algebraic factorization identity:

x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x)

We ruthlessly substitute this perfectly factored form into the denominator of our equation:

D (u) = \frac{3 ( x ^{2} + y ^{2} + z ^{2} - x y - y z - z x )}{( x + y + z ) ( x ^{2} + y ^{2} + z ^{2} - x y - y z - z x )}

The large polynomial bracket term elegantly cancels out from both the numerator and the denominator simultaneously, leaving the rigorously proven result:

D (u) = \frac{3}{x + y + z}

Part 2: Evaluating the Second-Order Operator $D^{2} (u)$

The notation $D^{2} (u)$ strictly means we must apply the differential operator $D$ again to the exact result we just derived. Mathematically, this is $D (D (u))$ :

D^{2} (u) = (\frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z}) (\frac{3}{x + y + z})

This demands that we calculate three new partial derivatives of the term $\frac{3}{x + y + z}$ and sum them up perfectly. We evaluate the derivative with respect to $x$ :

\frac{\partial}{\partial x} (3 (x + y + z)^{- 1}) = 3 (- 1) (x + y + z)^{- 2} \cdot \frac{\partial}{\partial x} (x + y + z) = \frac{- 3}{( x + y + z ) ^{2}} \cdot (1) = \frac{- 3}{( x + y + z ) ^{2}}

Due to the absolute continuing mathematical symmetry of the term with respect to $x$ , $y$ , and $z$ , differentiating with respect to $y$ and $z$ yields the exact identical mathematical result:

\frac{\partial}{\partial y} (\frac{3}{x + y + z}) = \frac{- 3}{( x + y + z ) ^{2}}

\frac{\partial}{\partial z} (\frac{3}{x + y + z}) = \frac{- 3}{( x + y + z ) ^{2}}

Finally, we rigorously sum these three newly derived identical fractional components to finalize the proof:

D^{2} (u) = \frac{- 3}{( x + y + z ) ^{2}} + \frac{- 3}{( x + y + z ) ^{2}} + \frac{- 3}{( x + y + z ) ^{2}} = \frac{- 9}{( x + y + z ) ^{2}}

Both required mathematical identities have now been completely and unequivocally verified.

Q19Mathematics I

Question

Answer

Part 1: Evaluating the First-Order Operator D(u)

Part 2: Evaluating the Second-Order Operator D2(u)

Part 1: Evaluating the First-Order Operator $D (u)$

Part 2: Evaluating the Second-Order Operator $D^{2} (u)$