Question

Solve the following differential equation ;
y'' - 2y' - 3y = e^2x

Answer 1

To solve the differential equation

\[
y'' - 2y' - 3y = e^{2x}
\]

we can use the method of undetermined coefficients. We first solve the associated homogeneous equation:

\[
y'' - 2y' - 3y = 0.
\]

The characteristic equation for this differential equation is:

\[
r^2 - 2r - 3 = 0.
\]

We can factor this as follows:

\[
(r - 3)(r + 1) = 0.
\]

Thus, the roots are:

\[
r = 3 \quad \text{and} \quad r = -1.
\]

The general solution of the homogeneous equation is:

\[
y_h = C_1 e^{3x} + C_2 e^{-x},
\]

where \(C_1\) and \(C_2\) are constants.

Next, we find a particular solution, \(y_p\), to the non-homogeneous equation. Since the right-hand side is \(e^{2x}\) and neither of the homogeneous solutions contains \(e^{2x}\), we can assume a particular solution of the form:

\[
y_p = Ae^{2x},
\]

where \(A\) is a constant to be determined.

Now we compute \(y_p'\) and \(y_p''\):

\[
y_p' = 2Ae^{2x},
\]
\[
y_p'' = 4Ae^{2x}.
\]

Substituting \(y_p\), \(y_p'\), and \(y_p''\) back into the original differential equation:

\[
4Ae^{2x} - 2(2Ae^{2x}) - 3(Ae^{2x}) = e^{2x}.
\]

This simplifies to:

\[
4Ae^{2x} - 4Ae^{2x} - 3Ae^{2x} = e^{2x}.
\]

Simplifying further:

\[
-3Ae^{2x} = e^{2x}.
\]

To find \(A\), we can set the coefficients equal:

\[
-3A = 1 \implies A = -\frac{1}{3}.
\]

So the particular solution is:

\[
y_p = -\frac{1}{3} e^{2x}.
\]

The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y = y_h + y_p = C_1 e^{3x} + C_2 e^{-x} - \frac{1}{3} e^{2x}.
\]

Thus, the final solution is:

\[
\boxed{y = C_1 e^{3x} + C_2 e^{-x} - \frac{1}{3} e^{2x}},
\]

where \(C_1\) and \(C_2\) are arbitrary constants.

Answer 2

Let me solve this second-order linear differential equation with constant coefficients step by step.

1) First, let's find the characteristic equation. For y'' - 2y' + 5y = 0
The characteristic equation is: r² - 2r + 5 = 0

2) Using the quadratic formula: r = [-(-2) ± √((-2)² - 4(1)(5))] / 2(1)
= [2 ± √(4 - 20)] / 2
= [2 ± √(-16)] / 2
= [2 ± 4i] / 2
= 1 ± 2i

3) Therefore, r₁ = 1 + 2i and r₂ = 1 - 2i

4) Since we have complex conjugate roots, the general solution will be of the form:
y = e^(ax)[c₁cos(bx) + c₂sin(bx)]
where a is the real part and b is the coefficient of i

5) In this case:
a = 1 (real part)
b = 2 (imaginary part)

Therefore, the general solution is:
y = e^x(c₁cos(2x) + c₂sin(2x))

where c₁ and c₂ are arbitrary constants determined by initial conditions (if any were given).

Answer 3

Your step-by-step solution is clear and accurate. Here’s a concise verification and recap of each step:

1. **Characteristic Equation**:
You correctly identified the characteristic equation for the differential equation \(y'' - 2y' + 5y = 0\):
\[
r^2 - 2r + 5 = 0.
\]

2. **Quadratic Formula**:
Applying the quadratic formula,
\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
gives:
\[
r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.
\]

3. **Roots**:
The roots are correctly computed:
\[
r_1 = 1 + 2i \quad \text{and} \quad r_2 = 1 - 2i.
\]

4. **General Solution Structure**:
With complex conjugate roots, the general solution is appropriately expressed as:
\[
y = e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)),
\]
where \(a\) is the real part and \(b\) is the coefficient of \(i\).

5. **Specific Values**:
Here, \(a = 1\) and \(b = 2\). Hence the general solution becomes:
\[
y = e^x(c_1 \cos(2x) + c_2 \sin(2x)).
\]

Overall, the final solution to the differential equation \(y'' - 2y' + 5y = 0\) is:
\[
\boxed{y = e^x(c_1 \cos(2x) + c_2 \sin(2x))},
\]
where \(c_1\) and \(c_2\) are arbitrary constants that can be determined using initial conditions. Excellent job!