Answer :
Sure, let's solve the differential equation step by step.
The given differential equation is:
[tex]\[ \frac{dy}{dx} = x + 2y + xy + 2 \][/tex]
We also have the initial condition:
[tex]\[ y(1) = 0 \][/tex]
### Step 1: Rewriting the Differential Equation
First, let's rewrite the differential equation for clarity:
[tex]\[ \frac{dy}{dx} = x(1 + y) + 2(1 + y) \][/tex]
### Step 2: Recognizing the Linear Form
This form doesn't easily suggest a common method directly. However, it can be helpful to recognize it in another form. Let’s rewrite it as:
[tex]\[ \frac{dy}{dx} = x + 2 + (2 + x)y \][/tex]
### Step 3: Applying the Method of Separation of Variables or Other Transformations
Instead of using separation of variables or partial fraction decomposition, which might not yield easily usable forms, let's think about an integrating factor here. However, the problem can become simplified by recognizing it has a standard solving technique.
### Step 4: Finding the Solution
Using the general solution techniques for first-order linear differential equations or through advanced solution methods, we rewrite the solution:
[tex]\[ y(x) = C \cdot e^{\int (2 + x) dx} - 1 \][/tex]
Let's compute that integrating factor exponent:
[tex]\[ \int (2 + x) dx = 2x + \frac{x^2}{2} \][/tex]
So,
[tex]\[ y(x) = C \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
### Step 5: Applying Initial Condition
To determine the constant [tex]\( C \)[/tex], we apply the initial condition [tex]\( y(1) = 0 \)[/tex]:
[tex]\[ 0 = C \cdot e^{2 \cdot 1 + \frac{1^2}{2}} - 1 \][/tex]
[tex]\[ 0 = C \cdot e^{2 + \frac{1}{2}} - 1 \][/tex]
[tex]\[ 1 = C \cdot e^{\frac{5}{2}} \][/tex]
[tex]\[ C = \frac{1}{e^{\frac{5}{2}}} = e^{-\frac{5}{2}} \][/tex]
Hence, substituting [tex]\( C \)[/tex] back into our solution:
[tex]\[ y(x) = e^{-\frac{5}{2}} \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
[tex]\[ y(x) = e^{-\frac{5}{2}} \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
Applying exponent properties:
[tex]\[ y(x) = e^{ -\frac{5}{2} + 2x+ \frac{x^2}{2}} - 1 \][/tex]
### Final Solution
[tex]\[ y(x) = e^{ 2x + \frac{x^2}{2} - \frac{5}{2}} - 1 \][/tex]
Therefore, the solution to the differential equation [tex]\(\frac{dy}{dx} = x + 2y + xy + 2\)[/tex] with the initial condition [tex]\(y(1) = 0\)[/tex] is:
[tex]\[ y(x) = e^{-5/2} \cdot e^{x(x/2 + 2)} - 1 \][/tex]
The given differential equation is:
[tex]\[ \frac{dy}{dx} = x + 2y + xy + 2 \][/tex]
We also have the initial condition:
[tex]\[ y(1) = 0 \][/tex]
### Step 1: Rewriting the Differential Equation
First, let's rewrite the differential equation for clarity:
[tex]\[ \frac{dy}{dx} = x(1 + y) + 2(1 + y) \][/tex]
### Step 2: Recognizing the Linear Form
This form doesn't easily suggest a common method directly. However, it can be helpful to recognize it in another form. Let’s rewrite it as:
[tex]\[ \frac{dy}{dx} = x + 2 + (2 + x)y \][/tex]
### Step 3: Applying the Method of Separation of Variables or Other Transformations
Instead of using separation of variables or partial fraction decomposition, which might not yield easily usable forms, let's think about an integrating factor here. However, the problem can become simplified by recognizing it has a standard solving technique.
### Step 4: Finding the Solution
Using the general solution techniques for first-order linear differential equations or through advanced solution methods, we rewrite the solution:
[tex]\[ y(x) = C \cdot e^{\int (2 + x) dx} - 1 \][/tex]
Let's compute that integrating factor exponent:
[tex]\[ \int (2 + x) dx = 2x + \frac{x^2}{2} \][/tex]
So,
[tex]\[ y(x) = C \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
### Step 5: Applying Initial Condition
To determine the constant [tex]\( C \)[/tex], we apply the initial condition [tex]\( y(1) = 0 \)[/tex]:
[tex]\[ 0 = C \cdot e^{2 \cdot 1 + \frac{1^2}{2}} - 1 \][/tex]
[tex]\[ 0 = C \cdot e^{2 + \frac{1}{2}} - 1 \][/tex]
[tex]\[ 1 = C \cdot e^{\frac{5}{2}} \][/tex]
[tex]\[ C = \frac{1}{e^{\frac{5}{2}}} = e^{-\frac{5}{2}} \][/tex]
Hence, substituting [tex]\( C \)[/tex] back into our solution:
[tex]\[ y(x) = e^{-\frac{5}{2}} \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
[tex]\[ y(x) = e^{-\frac{5}{2}} \cdot e^{2x + \frac{x^2}{2}} - 1 \][/tex]
Applying exponent properties:
[tex]\[ y(x) = e^{ -\frac{5}{2} + 2x+ \frac{x^2}{2}} - 1 \][/tex]
### Final Solution
[tex]\[ y(x) = e^{ 2x + \frac{x^2}{2} - \frac{5}{2}} - 1 \][/tex]
Therefore, the solution to the differential equation [tex]\(\frac{dy}{dx} = x + 2y + xy + 2\)[/tex] with the initial condition [tex]\(y(1) = 0\)[/tex] is:
[tex]\[ y(x) = e^{-5/2} \cdot e^{x(x/2 + 2)} - 1 \][/tex]
