Question

The solution of the differential equation \(\frac{dy}{dx} + \frac{y}{x} = x^2\) under the condition that y(1) = 1 is

• A. \(4xy = x^3 + 3\)
• B. \(4xy = x^4 + 3\)
• C. \(4xy = x^3 - 3\)
• D. \(4xy = x^4 - 3\)

Hint:

The integrating factor method is a standard procedure for first-order linear DEs. Remember the three key steps: find P(x) and Q(x), calculate the I.F., and then apply the solution formula \(y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \,dx + C\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to solve a first-order linear differential equation with a given initial condition (an Initial Value Problem).

Step 2: Key Formula or Approach:
The equation is in the linear form \(\frac{dy}{dx} + P(x)y = Q(x)\). We solve it using an integrating factor (I.F.).
1. Identify \(P(x)\) and \(Q(x)\).
2. Calculate the Integrating Factor: I.F. = \(e^{\int P(x)dx}\).
3. The general solution is given by: \(y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \,dx + C\).
4. Use the initial condition \(y(1)=1\) to find the value of the constant \(C\).

Step 3: Detailed Explanation:
The given equation is \(\frac{dy}{dx} + \frac{1}{x}y = x^2\).
1. Here, \(P(x) = \frac{1}{x}\) and \(Q(x) = x^2\).
2. Calculate the integrating factor:
\[ \text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\ln x} = x \quad (\text{for } x\gt 0) \]
3. Find the general solution:
\[ y \cdot x = \int x^2 \cdot x \,dx + C \]
\[ xy = \int x^3 \,dx + C \]
\[ xy = \frac{x^4}{4} + C \]
4. Apply the initial condition \(y(1) = 1\) (when \(x=1\), \(y=1\)):
\[ (1)(1) = \frac{(1)^4}{4} + C \]
\[ 1 = \frac{1}{4} + C \]
\[ C = 1 - \frac{1}{4} = \frac{3}{4} \]
Substitute the value of \(C\) back into the general solution:
\[ xy = \frac{x^4}{4} + \frac{3}{4} \]
To match the format of the options, multiply the entire equation by 4:
\[ 4xy = x^4 + 3 \]

Step 4: Final Answer:
The solution of the initial value problem is \(4xy = x^4 + 3\).