Question

If \(\frac{dy}{dx} + \frac{y}{x} = x^2\) then \(2y (2) - y (1) =\)

• A. \(\frac{11}{4}\)
• B. \(\frac{9}{4}\)
• C. \(\frac{15}{4}\)
• D. \(\frac{13}{4}\)

Answer 1

The Correct Option is C

We are given the linear differential equation: \[ \frac{dy}{dx} + \frac{y}{x} = x^2 \] 

This is a first-order linear differential equation.
Let’s solve it using the integrating factor (IF) method. 

Step 1: Identify the integrating factor (IF): \[ \text{IF} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x \] 

Step 2: Multiply the entire equation by the IF (x): \[ x \cdot \frac{dy}{dx} + y = x^3 \Rightarrow \frac{d}{dx}(x y) = x^3 \] 

Step 3: Integrate both sides: \[ \int \frac{d}{dx}(x y) \, dx = \int x^3 \, dx \Rightarrow x y = \frac{x^4}{4} + C \] 

Step 4: Solve for \(y\): \[ y = \frac{x^3}{4} + \frac{C}{x} \] 

Step 5: Compute \(2y(2) - y(1)\): \[ y(2) = \frac{8}{4} + \frac{C}{2} = 2 + \frac{C}{2} \] \[ y(1) = \frac{1}{4} + C \] \[ 2y(2) - y(1) = 2(2 + \frac{C}{2}) - \left(\frac{1}{4} + C\right) = 4 + C - \frac{1}{4} - C = 4 - \frac{1}{4} = \frac{15}{4} \] 

Final Answer: \(\frac{15}{4}\)

Answer 2

The given differential equation is: 

\[ \frac{dy}{dx} + \frac{1}{x} y = x^2 \]

This is a first-order linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = \frac{1}{x}\) and \(Q(x) = x^2\).

First, we find the integrating factor (I.F.):

\[ I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} \]

Assuming \(x > 0\) since we are interested in \(y(1)\) and \(y(2)\), the integrating factor is:

\[ I.F. = e^{\ln x} = \mathbf{x} \]

Next, multiply the entire differential equation by the integrating factor \(x\):

\[ x \left( \frac{dy}{dx} + \frac{1}{x} y \right) = x(x^2) \]

\[ x \frac{dy}{dx} + y = x^3 \]

The left side of the equation is the derivative of the product of \(y\) and the integrating factor:

\[ \frac{d}{dx}(x \cdot y) = x^3 \]

Integrate both sides with respect to \(x\):

\[ \int \frac{d}{dx}(xy) \, dx = \int x^3 \, dx \]

\[ xy = \frac{x^4}{4} + C \]

where \(C\) is the constant of integration.

The general solution for \(y(x)\) is:

\[ \mathbf{y(x) = \frac{x^3}{4} + \frac{C}{x}} \]

Now we need to find the values of \(y(1)\) and \(y(2)\):

\[ y(1) = \frac{(1)^3}{4} + \frac{C}{1} = \mathbf{\frac{1}{4} + C} \]

\[ y(2) = \frac{(2)^3}{4} + \frac{C}{2} = \frac{8}{4} + \frac{C}{2} = \mathbf{2 + \frac{C}{2}} \]

Finally, we compute the required expression \(2y(2) - y(1)\):

\[ 2y(2) - y(1) = 2\left(2 + \frac{C}{2}\right) - \left(\frac{1}{4} + C\right) \]

\[ = (4 + C) - \left(\frac{1}{4} + C\right) \]

\[ = 4 + C - \frac{1}{4} - C \]

\[ = 4 - \frac{1}{4} \]

\[ = \frac{16}{4} - \frac{1}{4} \]

\[ = \mathbf{\frac{15}{4}} \]

Comparing this with the given options, the correct option is:

\(\frac{15}{4}\)