Solve the following differential equation $$ y^{\prime \prime \prime}-3 y^{\prime \prime}=x e^{2 x}+6 $$ Select one: a.. $$ y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-e^{2 x}-1 $$ b. $$ y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-x e^{2 x}-2 e^{2 x}-3 $$ c. $$ y_{G . S}=c_{1}+c_{2}+c_{3} e^{3 x}-0.25 x e^{2 x}+1 $$ d. $$ y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-x e^{2 x}-e^{2 x} $$ e. $$ y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-0.25 x e^{2 x}-1 $$

Question

UpStudy AI · Accepted Answer

To solve the differential equation

$$
y^{\prime \prime \prime}-3 y^{\prime \prime}=x e^{2 x}+6,
$$

we will first find the general solution $$ y_G $$ and then identify the particular solution $$ y_P $$.

### Step 1: Find the Complementary Solution $$ y_C $$

The complementary solution is found by solving the associated homogeneous equation:

$$
y^{\prime \prime \prime} - 3 y^{\prime \prime} = 0.
$$

The characteristic equation is:

$$
r^3 - 3r^2 = 0.
$$

Factoring gives:

$$
r^2(r - 3) = 0.
$$

This results in roots $$ r = 0, 0, 3 $$. Therefore, the complementary solution is:

$$
y_C = c_1 + c_2 x + c_3 e^{3x}.
$$

### Step 2: Find the Particular Solution $$ y_P $$

Next, we need to find a particular solution for the non-homogeneous part $$ x e^{2x} + 6 $$.

For the term $$ x e^{2x} $$, we will use the method of undetermined coefficients. Since $$ e^{2x} $$ is not a solution of the homogeneous equation, we can try a particular solution of the form:

$$
y_P = A x e^{2x} + B e^{2x} + C,
$$

where $$ A, B, $$ and $$ C $$ are constants to be determined.

### Step 3: Substitute $$ y_P $$ into the Differential Equation

We will compute the derivatives of $$ y_P $$ and substitute them back into the original differential equation.

1. First derivative:
   $$
   y_P' = A e^{2x} + 2A x e^{2x} + 2B e^{2x}.
   $$

2. Second derivative:
   $$
   y_P'' = 2A e^{2x} + 2A e^{2x} + 4A x e^{2x} + 4B e^{2x} = 4A x e^{2x} + 4(A + B)e^{2x}.
   $$

3. Third derivative:
   $$
   y_P''' = 4A e^{2x} + 4A e^{2x} + 8A x e^{2x} + 8(A + B)e^{2x} = 8A x e^{2x} + 12(A + B)e^{2x}.
   $$

Now substituting $$ y_P, y_P'', $$ and $$ y_P''' $$ into the original equation:

$$
(8A x e^{2x} + 12(A + B)e^{2x}) - 3(4A x e^{2x} + 4(A + B)e^{2x}) = x e^{2x} + 6.
$$

### Step 4: Collect Like Terms

This simplifies to:

$$
(8A - 12A)x e^{2x} + (12A + 12B - 12A - 12B)e^{2x} = x e^{2x} + 6.
$$

This gives us:

$$
-4A x e^{2x} = x e^{2x} \quad \text{and} \quad 0 = 6.
$$

From $$ -4A = 1 $$, we find $$ A = -\frac{1}{4} $$.

For the constant term, we can set $$ C = -1 $$ to satisfy the equation.

### Step 5: Combine Solutions

Thus, the particular solution is:

$$
y_P = -\frac{1}{4} x e^{2x} - 1.
$$

### Final General Solution

Combining the complementary and particular solutions, we have:

$$
y = c_1 + c_2 x + c_3 e^{3x} - \frac{1}{4} x e^{2x} - 1.
$$

### Step 6: Match with Given Options

Now, we can compare this with the provided options. The correct form is:

$$
y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-0.25 x e^{2 x}-1.
$$

Thus, the correct answer is:

**e. $$ y_{G . S}=c_{1}+c_{2} x+c_{3} e^{3 x}-0.25 x e^{2 x}-1 $$**.
The correct general solution is $$ y_{G.S} = c_{1} + c_{2} x + c_{3} e^{3x} - 0.25 x e^{2x} - 1 $$.

Solve the following differential equation

Select one:
a…
b.
c.
d.
e.

Upstudy AI Solution

Answer

Solution

Step 1: Find the Complementary Solution

Step 2: Find the Particular Solution

Step 3: Substitute into the Differential Equation

Step 4: Collect Like Terms

Step 5: Combine Solutions

Final General Solution

Step 6: Match with Given Options

Beyond the Answer

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