If \(y = y(x)\) is the solution of the differential equation \(\displaystyle\frac{dy}{dx} - \frac{3y}{x} = x^3\), \(x \gt 0\), with \(y(1) = 3\), then \(y(2)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 32
This is a linear equation with \(P = -\frac{3}{x}\) and \(Q = x^3\). The integrating factor is
\( \displaystyle e^{-3\log x} = \frac{1}{x^3} \)
So
\( \displaystyle \frac{y}{x^3} = \int x^3 \cdot \frac{1}{x^3}\,dx = x + C \)
\( y = x^4 + Cx^3 \)
Using \(y(1) = 3\):
\( 3 = 1 + C \Rightarrow C = 2 \)
So \(y = x^4 + 2x^3\), and
\( y(2) = 16 + 16 = 32 \)
Why the other options are incorrect
Option (1) 18 — Incorrect. This takes \(y = x^4 + C\), giving \(16 + 2\); the constant must also be multiplied by \(x^3\), which gives \(y = x^4 + Cx^3\).
Option (2) 40 — Incorrect. This takes \(C = 3\), the value of y at 1, but \(y(1) = 1 + C\) gives \(C = 2\).
Option (3) 16 — Incorrect. This is \(x^4\) at \(x = 2\); the term \(Cx^3\) is dropped, and \(y = x^4\) does not give \(y(1) = 3\).