Using the chain rule, the derivative of e^(ax) is ae^(ax). Applying this, the derivative of 3e²x is 3 · 2e²x = 6e²x, and the derivative of 2e³x is 2 · 3e³x = 6e³x.

Now we differentiate the first derivative, 6e²x + 6e³x, to find the second derivative of y with respect to x.

Again, applying the chain rule, the derivative of 6e²x is 6 · 2e²x = 12e²x, and the derivative of 6e³x is 6 · 3e³x = 18e³x.

Now we substitute the expressions for y, (dy)/(dx), and (d²y)/(dx²) into the given equation (d² y)/(d x²)-5 (d y)/(d x)+6 y=0.

Distribute the coefficients and expand the terms. We will then group like terms (e²x terms and e³x terms) to simplify the expression.

Combine the coefficients for e²x and e³x terms. We find that (12 - 30 + 18) = 0 and (18 - 30 + 12) = 0, which proves the equation is equal to zero.

If $y=3 e^{2 x}+2 e^{3 x}$ , prove that $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$ .

Answer: The given equation $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$ is proven to be true.

Step-by-step solution

Step 1: Find the first derivative

To prove the given equation, we first need to find the first and second derivatives of $y$ with respect to $x$ . We start by differentiating the given function $y = 3e^{2x} + 2e^{3x}$ once.

Step 2: Calculate the first derivative

Using the chain rule, the derivative of $e^{ax}$ is $ae^{ax}$ . Applying this, the derivative of $3e^{2x}$ is $3 \cdot 2e^{2x} = 6e^{2x}$ , and the derivative of $2e^{3x}$ is $2 \cdot 3e^{3x} = 6e^{3x}$ .

Step 3: Find the second derivative

Now we differentiate the first derivative, $6e^{2x} + 6e^{3x}$ , to find the second derivative of $y$ with respect to $x$ .

Step 4: Calculate the second derivative

Again, applying the chain rule, the derivative of $6e^{2x}$ is $6 \cdot 2e^{2x} = 12e^{2x}$ , and the derivative of $6e^{3x}$ is $6 \cdot 3e^{3x} = 18e^{3x}$ .

Step 5: Substitute into the equation

Now we substitute the expressions for $y$ , $\frac{dy}{dx}$ , and $\frac{d^2y}{dx^2}$ into the given equation $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0$ .

Step 6: Simplify the expression

Distribute the coefficients and expand the terms. We will then group like terms ( $e^{2x}$ terms and $e^{3x}$ terms) to simplify the expression.

Step 7: Combine like terms

Combine the coefficients for $e^{2x}$ and $e^{3x}$ terms. We find that $(12 - 30 + 18) = 0$ and $(18 - 30 + 12) = 0$ , which proves the equation is equal to zero.

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If y=3e2x+2e3xy=3 e^{2 x}+2 e^{3 x}y=3e2x+2e3x, prove that d2ydx2−5dydx+6y=0\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0dx2d2y​−5dxdy​+6y=0.