b) Find the homogeneous solution to $$ y^{\prime \prime}+4 y=f(t) $$. For each $$ f(t) $$, write down the particular solution $$ y_{p} $$ that will solve the problem; i. $$ f(t)=4 t^{2}+1 $$, ii. $$ f(t)=3 \sin (2 t) $$ iii. $$ f(t)=e^{-2 t}-2 t $$

Question

b) Find the homogeneous solution to $$ y^{\prime \prime}+4 y=f(t) $$. For each $$ f(t) $$, write down the particular solution $$ y_{p} $$ that will solve the problem; 	i. $$ f(t)=4 t^{2}+1 $$, 	ii. $$ f(t)=3 \sin (2 t) $$ 	iii. $$ f(t)=e^{-2 t}-2 t $$

UpStudy AI · Accepted Answer

The homogeneous equation is

$$
y'' + 4y = 0.
$$

Its characteristic equation is

$$
r^2 + 4 = 0 \quad \Longrightarrow \quad r = \pm 2i.
$$

Thus, the homogeneous solution is

$$
y_h = C_1\cos(2t) + C_2\sin(2t).
$$

---

**(i) For $$ f(t)=4t^2+1 $$:**

We try a particular solution of the form

$$
y_p = At^2 + Bt + C.
$$

Then

$$
y_p' = 2At + B, \qquad y_p'' = 2A.
$$

Substitute into the differential equation:

$$
2A + 4(At^2 + Bt + C) = 4t^2 + 1.
$$

Collecting like terms we get

$$
4At^2 + 4Bt + (2A+4C) = 4t^2 + 0 \cdot t + 1.
$$

Equate coefficients:

- Coefficient of $$ t^2 $$: $$ 4A = 4 $$ ⟹ $$ A = 1 $$.
- Coefficient of $$ t $$: $$ 4B = 0 $$ ⟹ $$ B = 0 $$.
- Constant term: $$ 2A + 4C = 1 $$. Since $$ A=1 $$, we have

$$
  2 + 4C = 1 \quad \Longrightarrow \quad 4C = -1 \quad \Longrightarrow \quad C = -\frac{1}{4}.
  $$

Thus,

$$
y_p = t^2 -\frac{1}{4}.
$$

---

**(ii) For $$ f(t)=3\sin(2t) $$:**

Since both $$\cos(2t)$$ and $$\sin(2t)$$ are part of the homogeneous solution, we use the method of undetermined coefficients with a factor of $$ t $$. Let

$$
y_p = t\Bigl( A\cos(2t) + B\sin(2t) \Bigr).
$$

Differentiate:

$$
\begin{aligned}
y_p &= t\Bigl( A\cos(2t) + B\sin(2t) \Bigr), \
y_p' &= A\cos(2t) + B\sin(2t) + t\Bigl(-2A\sin(2t) + 2B\cos(2t) \Bigr), \
y_p'' &= -2A\sin(2t) + 2B\cos(2t) \
&\quad {} + \Bigl[-2A\sin(2t) + 2B\cos(2t)\Bigr] + t\Bigl(-4A\cos(2t)-4B\sin(2t)\Bigr) \
&= -4A\sin(2t) + 4B\cos(2t) - 4t\Bigl[A\cos(2t)+B\sin(2t)\Bigr].
\end{aligned}
$$

Now, substitute $$y_p$$ and $$y_p''$$ into $$y''+4y$$:

$$
\begin{aligned}
y_p''+4y_p &= \Bigl[-4A\sin(2t)+4B\cos(2t)-4t\Bigl(A\cos(2t)+B\sin(2t)\Bigr)\Bigr] \
&\quad {} + 4t\Bigl(A\cos(2t)+B\sin(2t)\Bigr) \
&= -4A\sin(2t)+4B\cos(2t).
\end{aligned}
$$

We need

$$
-4A\sin(2t)+4B\cos(2t)=3\sin(2t).
$$

Equate coefficients:

- For $$\sin(2t)$$: $$-4A = 3 $$ ⟹ $$ A = -\frac{3}{4} $$.
- For $$\cos(2t)$$: $$ 4B = 0 $$ ⟹ $$ B = 0 $$.

Thus,

$$
y_p = -\frac{3}{4}t\cos(2t).
$$

---

**(iii) For $$ f(t)=e^{-2t}-2t $$:**

We split the forcing function and find a particular solution for each part.

*For the $$ e^{-2t} $$ part:*

Assume

$$
y_{p1} = Ae^{-2t}.
$$

Then

$$
y_{p1}' = -2Ae^{-2t}, \qquad y_{p1}'' = 4Ae^{-2t}.
$$

Substitute into the equation:

$$
4Ae^{-2t} + 4Ae^{-2t} = 8Ae^{-2t} = e^{-2t}.
$$

Thus,

$$
8A = 1 \quad \Longrightarrow \quad A = \frac{1}{8}.
$$

So,

$$
y_{p1} = \frac{1}{8}e^{-2t}.
$$

*For the $$-2t$$ part:*

Assume

$$
y_{p2} = Bt + C.
$$

Then

$$
y_{p2}' = B, \qquad y_{p2}'' = 0.
$$

Substitute into the equation:

$$
0 + 4(Bt+C) = 4Bt+4C = -2t.
$$

Equate coefficients:

- Coefficient of $$ t $$: $$ 4B = -2 $$ ⟹ $$ B = -\frac{1}{2} $$.
- Constant term: $$ 4C = 0 $$ ⟹ $$ C = 0 $$.

So,

$$
y_{p2} = -\frac{1}{2}t.
$$

The overall particular solution is the sum:

$$
y_p = y_{p1} + y_{p2} = \frac{1}{8}e^{-2t} - \frac{1}{2}t.
$$

---

Final answers:

- Homogeneous solution:
  
  $$
  y_h = C_1\cos(2t) + C_2\sin(2t).
  $$

- (i) When $$\displaystyle f(t)=4t^2+1$$:
  
  $$
  y_p = t^2 -\frac{1}{4}.
  $$

- (ii) When $$\displaystyle f(t)=3\sin(2t)$$:
  
  $$
  y_p = -\frac{3}{4}t\cos(2t).
  $$

- (iii) When $$\displaystyle f(t)=e^{-2t}-2t$$:
  
  $$
  y_p = \frac{1}{8}e^{-2t} - \frac{1}{2}t.
  $$
Homogeneous solution: $$ y_h = C_1\cos(2t) + C_2\sin(2t) $$. **(i) For $$ f(t)=4t^2+1 $$:** Particular solution: $$ y_p = t^2 -\frac{1}{4} $$. **(ii) For $$ f(t)=3\sin(2t) $$:** Particular solution: $$ y_p = -\frac{3}{4}t\cos(2t) $$. **(iii) For $$ f(t)=e^{-2t}-2t $$:** Particular solution: $$ y_p = \frac{1}{8}e^{-2t} - \frac{1}{2}t $$.

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i. ,
ii.
iii.

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