Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (I.F.): For a first-order linear differential equation, the integrating factor is $I.F. = e^{\int P(x) dx}$.
• General Solution: The general solution to a first-order linear differential equation is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.

Step-by-Step Solution

Step 1: Rewrite the given differential equation in standard form.

We are given the differential equation $x \frac{dy}{dx} + y = bx^4$. To put it in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide both sides by $x$: $\frac{dy}{dx} + \frac{1}{x}y = bx^3$ This is a first-order linear differential equation with $P(x) = \frac{1}{x}$ and $Q(x) = bx^3$. We are doing this to identify P(x) and Q(x) so we can find the integrating factor and solve the equation.

Step 2: Calculate the integrating factor (I.F.).

The integrating factor is given by $I.F. = e^{\int P(x) dx}$. In this case, $P(x) = \frac{1}{x}$, so $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$ We take the positive value, $x$, since it is an integrating factor.

Step 3: Find the general solution of the differential equation.

The general solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. Substituting $I.F. = x$ and $Q(x) = bx^3$, we have $yx = \int (bx^3)(x) dx + C = \int bx^4 dx + C = b\frac{x^5}{5} + C$ Thus, the general solution is $y = \frac{bx^4}{5} + \frac{C}{x}$

Step 4: Use the initial condition to find the particular solution.

We are given that the curve passes through the point (1, 2), so $f(1) = 2$. Substituting $x = 1$ and $y = 2$ into the general solution, we get $2 = \frac{b(1)^4}{5} + \frac{C}{1} = \frac{b}{5} + C$ So, $C = 2 - \frac{b}{5}$. Substituting this back into the general solution, we obtain the particular solution: $y = \frac{bx^4}{5} + \frac{2 - \frac{b}{5}}{x} = \frac{bx^4}{5} + \frac{2}{x} - \frac{b}{5x}$

Step 5: Use the given integral to solve for b.

We are given that $\int_1^2 f(x) dx = \frac{62}{5}$. Substituting our expression for $f(x) = y$, we have $\int_1^2 \left(\frac{bx^4}{5} + \frac{2}{x} - \frac{b}{5x}\right) dx = \frac{62}{5}$ Now we evaluate the integral: $\left[\frac{bx^5}{25} + 2\ln|x| - \frac{b}{5}\ln|x|\right]_1^2 = \frac{62}{5}$ $\left(\frac{b(2^5)}{25} + 2\ln 2 - \frac{b}{5}\ln 2\right) - \left(\frac{b(1^5)}{25} + 2\ln 1 - \frac{b}{5}\ln 1\right) = \frac{62}{5}$ $\frac{32b}{25} + 2\ln 2 - \frac{b}{5}\ln 2 - \frac{b}{25} - 0 + 0 = \frac{62}{5}$ $\frac{31b}{25} + \left(2 - \frac{b}{5}\right)\ln 2 = \frac{62}{5}$ $\frac{31b}{25} + 2\ln 2 - \frac{b}{5}\ln 2 = \frac{62}{5}$ Multiply by 25: $31b + 50\ln 2 - 5b\ln 2 = 310$ $31b - 5b\ln 2 = 310 - 50\ln 2$ $b(31 - 5\ln 2) = 10(31 - 5\ln 2)$ $b = 10$

Step 6: Re-evaluate with Correct Answer Since the given answer is $b = \frac{31}{5}$, there must be an error above. Let's carefully re-examine the calculations. We had: $\int_1^2 \left(\frac{bx^4}{5} + \frac{2}{x} - \frac{b}{5x}\right) dx = \frac{62}{5}$ $\left[\frac{bx^5}{25} + 2\ln x - \frac{b}{5}\ln x\right]_1^2 = \frac{62}{5}$ $\left(\frac{32b}{25} + 2\ln 2 - \frac{b}{5}\ln 2\right) - \left(\frac{b}{25} + 0 - 0\right) = \frac{62}{5}$ $\frac{31b}{25} + \left(2 - \frac{b}{5}\right)\ln 2 = \frac{62}{5}$ It seems the error is that we must have $2 - \frac{b}{5} = 0$, so that the $\ln 2$ terms disappear. However, this would require $b = 10$, which isn't the correct answer. So, let's assume we made a mistake in simplifying, and continue. $\frac{31b}{25} + 2\ln 2 - \frac{b\ln 2}{5} = \frac{62}{5}$ $\frac{31b}{25} - \frac{b\ln 2}{5} = \frac{62}{5} - 2\ln 2$ $b\left(\frac{31}{25} - \frac{\ln 2}{5}\right) = \frac{62}{5} - 2\ln 2$ $b\left(\frac{31 - 5\ln 2}{25}\right) = \frac{62 - 10\ln 2}{5}$ $b = \frac{25(62 - 10\ln 2)}{5(31 - 5\ln 2)} = \frac{5(62 - 10\ln 2)}{31 - 5\ln 2} = \frac{10(31 - 5\ln 2)}{31 - 5\ln 2} = 10$ Still getting $b = 10$.

Let's go back to Step 5, and assume $b = \frac{31}{5}$. $\int_1^2 (\frac{bx^4}{5} + \frac{2}{x} - \frac{b}{5x}) dx = \frac{62}{5}$ $\int_1^2 (\frac{31x^4}{25} + \frac{2}{x} - \frac{31}{25x}) dx = \frac{62}{5}$ $[\frac{31x^5}{125} + 2\ln x - \frac{31}{25}\ln x]_1^2 = \frac{62}{5}$ $(\frac{31(32)}{125} + 2\ln 2 - \frac{31}{25}\ln 2) - (\frac{31}{125} + 0 - 0) = \frac{62}{5}$ $\frac{992}{125} - \frac{31}{125} + 2\ln 2 - \frac{31}{25}\ln 2 = \frac{62}{5}$ $\frac{961}{125} + \frac{50 - 31}{25}\ln 2 = \frac{62}{5}$ $\frac{961}{125} + \frac{19}{25}\ln 2 = \frac{62}{5} = \frac{1550}{125}$ $\frac{19}{25}\ln 2 = \frac{1550 - 961}{125} = \frac{589}{125}$ $\ln 2 = \frac{589}{125} \cdot \frac{25}{19} = \frac{31}{5}$ This is incorrect, as $\ln 2 \neq \frac{31}{5}$.

Going back to step 3, if we have $y = \frac{bx^4}{5} + \frac{C}{x}$. And $y(1) = 2$, so $2 = \frac{b}{5} + C$, and $C = 2 - \frac{b}{5}$. Then $\int_1^2 (\frac{bx^4}{5} + \frac{2}{x} - \frac{b}{5x}) dx = \frac{62}{5}$. $\frac{bx^5}{25} + 2\ln x - \frac{b\ln x}{5}|_1^2 = \frac{62}{5}$ $(\frac{32b}{25} + 2\ln 2 - \frac{b\ln 2}{5}) - (\frac{b}{25}) = \frac{62}{5}$ $\frac{31b}{25} + 2\ln 2 - \frac{b\ln 2}{5} = \frac{62}{5}$ $31b + 50\ln 2 - 5b\ln 2 = 310$ $b(31 - 5\ln 2) = 310 - 50\ln 2$ $b = \frac{310 - 50\ln 2}{31 - 5\ln 2} = 10$

There seems to be an error in the question or the answer key. Given the problem constraints, the correct answer is $b = 10$.

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always remember to add the constant of integration, $C$, when evaluating indefinite integrals.
• Incorrectly Applying the Initial Condition: Ensure that you substitute the initial condition correctly to find the value of the constant $C$.
• Algebra Errors: Double-check your algebraic manipulations, especially when dealing with fractions and logarithms.

Summary

We solved the first-order linear differential equation using an integrating factor and applied the initial condition to find the particular solution. We then used the given definite integral to solve for the value of $b$. However, the calculated value $b=10$ does not match the stated correct answer of $\frac{31}{5}$. Based on the problem, the correct answer should be 10. There may be an error in the problem statement or the provided answer options.

Final Answer

The final answer is \boxed{10}, which does not correspond to any of the options. The closest option is (A), but the correct answer is 10.