If `y` is a function of `x` then `(d^2y)/(dx^2)+y \ dy/dx=0.` If `x` is a function of `y` then the equation becomes

Options

A. `(d^(2)x)/(dy^(2))-x(dx)/(dy)=0`
B. `(d^(2)x)/(dy^(2))+y((dx)/(dy))^(2)=0`
C. `(d^(2)x)/(dy^(2))-y((dx)/(dy))^(2)=0`
D. `(d^(2)x)/(dy^(2))-x((dx)/(dy))^(2)=0`

Answer

To solve the given problem, we start with the differential equation:

\[ \frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0 \]

We need to express this equation in terms of \( x \) as a function of \( y \).

### Step 1: Rewrite \(\frac{dy}{dx}\)

We know that:

\[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \]

### Step 2: Rewrite \(\frac{d^2y}{dx^2}\)

Using the chain rule, we can express \(\frac{d^2y}{dx^2}\) in terms of \(\frac{dx}{dy}\):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right) \]

Using the quotient rule, we have:

\[ \frac{d^2y}{dx^2} = -\frac{1}{\left(\frac{dx}{dy}\right)^2} \cdot \frac{d^2x}{dy^2} \]

### Step 3: Substitute into the original equation

Now we substitute this expression back into the original equation:

\[ -\frac{1}{\left(\frac{dx}{dy}\right)^2} \cdot \frac{d^2x}{dy^2} + y \cdot \frac{1}{\frac{dx}{dy}} = 0 \]

### Step 4: Multiply through by \(\left(\frac{dx}{dy}\right)^2\)

To eliminate the fraction, we multiply through by \(\left(\frac{dx}{dy}\right)^2\):

\[ -\frac{d^2x}{dy^2} + y \cdot \frac{dx}{dy} = 0 \]

### Step 5: Rearranging the equation

Rearranging gives us:

\[ \frac{d^2x}{dy^2} - y \cdot \frac{dx}{dy} = 0 \]

This is the required form of the differential equation when \( x \) is a function of \( y \).

### Final Result

Thus, the equation becomes:

\[ \frac{d^2x}{dy^2} - y \cdot \frac{dx}{dy} = 0 \]

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