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: Key Idea:
For an equation of the form $y''-2py'+(p^2+q^2)y=0$, the general solution is known to be $y=e^{px}(A\cos qx+B\sin qx)$.
This is a standard result for linear ODEs with constant coefficients when the auxiliary equation has complex roots.

Step 2: Form and solve the auxiliary equation:
For $\dfrac{d^2y}{dx^2}-6\dfrac{dy}{dx}+25y=0$, replace derivatives by powers of m.
\[ m^2-6m+25=0 \]
Solve using the quadratic formula.
\[ m=\frac{6\pm\sqrt{36-100}}{2}=\frac{6\pm\sqrt{-64}}{2}=3\pm4i \]

Step 3: Write the general solution from the roots:
The roots are complex conjugates $p\pm iq$ with $p=3$ and $q=4$.
By the standard rule, the general solution is $y=e^{3x}(A\cos4x+B\sin4x)$, using two independent constants A and B.
This is exactly the given function $y=Ae^{3x}\cos4x+Be^{3x}\sin4x$.

Final Answer:
Since the given y matches the general solution built from the auxiliary equation roots, it solves the differential equation. \[ \boxed{y=Ae^{3x}\cos4x+Be^{3x}\sin4x \text{ satisfies } y''-6y'+25y=0} \]