Question

Prove that the function \(y=Ae^{3x}\cos4x+Be^{3x}\sin4x\), where \(A\), \(B\) are arbitrary constants, is the solution of the differential equation \(\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+25y=0\).

Hint:

Differentiate y twice and substitute into the equation; check the coefficients of cos4x and sin4x cancel to zero.

Answer 1

Step 1: Understanding the Concept:
The function \(y=Ae^{3x}\cos4x+Be^{3x}\sin4x\) is given.
To prove it is a solution, differentiate y twice and put the values of y, dy/dx and d^2y/dx^2 into the left side of the equation.
If the left side reduces to zero, the function is a solution.

Step 2: Find dy/dx:
Differentiate each term using the product rule.
\[ \frac{dy}{dx}=e^{3x}\left[(3A+4B)\cos4x+(3B-4A)\sin4x\right] \]

Step 3: Find d^2y/dx^2:
Differentiate dy/dx again using the product rule on the new expression.
\[ \frac{d^2y}{dx^2}=e^{3x}\left[(-7A+24B)\cos4x+(-24A-7B)\sin4x\right] \]

Step 4: Substitute into the equation:
Put d^2y/dx^2, dy/dx and y into \(\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+25y\).
Collect the coefficient of \(\cos4x\): \((-7A+24B)-6(3A+4B)+25A=(-7-18+25)A+(24-24)B=0\).
Collect the coefficient of \(\sin4x\): \((-24A-7B)-6(3B-4A)+25B=(-24+24)A+(-7-18+25)B=0\).
\[ \frac{d^2y}{dx^2}-6\frac{dy}{dx}+25y=0 \]

Final Answer:
The left side becomes zero for all A and B, so y satisfies the equation. \[ \boxed{y=Ae^{3x}\cos4x+Be^{3x}\sin4x \text{ is a solution of the equation}} \]