Given `y=y(x)`passing through (1,2) such that `xdy/dx +y = bx^4` then find b if `int_1^(2) f(x)dx=62/5`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Rewrite the Differential Equation
The differential equation given is:
\[ x \frac{dy}{dx} + y = b x^4 \]
We can rearrange this into a standard linear form:
\[ \frac{dy}{dx} + \frac{y}{x} = b x^3 \]

### Step 2: Identify the Integrating Factor
The integrating factor \( \mu(x) \) for the equation \( \frac{dy}{dx} + P(x)y = Q(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = x \]

### Step 3: Multiply the Equation by the Integrating Factor
Multiplying the entire equation by \( x \):
\[ x \frac{dy}{dx} + y = b x^4 \]
This simplifies to:
\[ \frac{d}{dx}(xy) = b x^4 \]

### Step 4: Integrate Both Sides
Integrating both sides:
\[ xy = \int b x^4 \, dx = \frac{b}{5} x^5 + C \]
Thus, we have:
\[ y = \frac{b}{5} x^4 + \frac{C}{x} \]

### Step 5: Use the Initial Condition
The curve passes through the point (1, 2):
\[ 2 = \frac{b}{5} \cdot 1^4 + C \implies 2 = \frac{b}{5} + C \]
From this, we can express \( C \):
\[ C = 2 - \frac{b}{5} \]

### Step 6: Substitute \( C \) Back into the Equation for \( y \)
Substituting \( C \) back into the equation for \( y \):
\[ y = \frac{b}{5} x^4 + \frac{2 - \frac{b}{5}}{x} \]

### Step 7: Set Up the Integral
We need to evaluate the integral:
\[ \int_1^2 y \, dx = \int_1^2 \left( \frac{b}{5} x^4 + \frac{2 - \frac{b}{5}}{x} \right) dx \]
This can be split into two parts:
\[ \int_1^2 \frac{b}{5} x^4 \, dx + \int_1^2 \left( 2 - \frac{b}{5} \right) \frac{1}{x} \, dx \]

### Step 8: Calculate Each Integral
1. For the first integral:
\[ \int_1^2 \frac{b}{5} x^4 \, dx = \frac{b}{5} \cdot \left[ \frac{x^5}{5} \right]_1^2 = \frac{b}{5} \cdot \left( \frac{32}{5} - \frac{1}{5} \right) = \frac{b}{5} \cdot \frac{31}{5} = \frac{31b}{25} \]

2. For the second integral:
\[ \int_1^2 \left( 2 - \frac{b}{5} \right) \frac{1}{x} \, dx = \left( 2 - \frac{b}{5} \right) \cdot \left[ \ln x \right]_1^2 = \left( 2 - \frac{b}{5} \right) \cdot \ln 2 \]

### Step 9: Combine the Results
Combining both results, we have:
\[ \int_1^2 y \, dx = \frac{31b}{25} + \left( 2 - \frac{b}{5} \right) \ln 2 \]
Setting this equal to \( \frac{62}{5} \):
\[ \frac{31b}{25} + \left( 2 - \frac{b}{5} \right) \ln 2 = \frac{62}{5} \]

### Step 10: Solve for \( b \)
Now we need to solve this equation for \( b \):
1. Rearranging gives:
\[ \frac{31b}{25} - \frac{b}{5} \ln 2 + 2 \ln 2 = \frac{62}{5} \]
2. Multiply through by 25 to eliminate the fraction:
\[ 31b - 5b \ln 2 + 50 \ln 2 = 310 \]
3. Rearranging gives:
\[ b(31 - 5 \ln 2) = 310 - 50 \ln 2 \]
4. Finally, solving for \( b \):
\[ b = \frac{310 - 50 \ln 2}{31 - 5 \ln 2} \]

### Step 11: Approximate \( b \)
Using \( \ln 2 \approx 0.693 \):
\[ b \approx \frac{310 - 50 \cdot 0.693}{31 - 5 \cdot 0.693} \approx \frac{310 - 34.65}{31 - 3.465} \approx \frac{275.35}{27.535} \approx 10 \]

Thus, the value of \( b \) is approximately \( 10 \).