Question

Consider differential equation \(\dfrac{dy}{dx}+xy=x\) with the condition as \(y=0\) at \(x=0\). The value of \(y\) at \(x=1.0\) is ______ (rounded off to two decimal places).

Hint:

Use an integrating factor \(e^{x^2/2}\) on \(\dfrac{dy}{dx}+xy=x\), or substitute \(v=1-y\) to get a separable equation; either way \(y=1-e^{-x^2/2}\).

Answer 1

Correct Answer: 0.39

Step 1: Identify the type of equation.
The equation \(\dfrac{dy}{dx}+xy=x\) is a first order linear differential equation of the standard form \(\dfrac{dy}{dx}+P(x)y=Q(x)\), with \(P(x)=x\) and \(Q(x)=x\).
A linear first order equation like this is solved using an integrating factor.

Step 2: Find the integrating factor.
The integrating factor is \(\mu(x)=e^{\int P(x)\,dx}=e^{\int x\,dx}=e^{x^2/2}\).
Multiplying both sides of the equation by \(\mu(x)\) turns the left side into the derivative of a product:
\[ \frac{d}{dx}\Big(y\,e^{x^2/2}\Big)=x\,e^{x^2/2} \]
Step 3: Integrate both sides.
\[ y\,e^{x^2/2}=\int x\,e^{x^2/2}\,dx \] For the right side, let \(u=x^2/2\), so \(du=x\,dx\), and the integral becomes \(\int e^{u}\,du=e^{u}=e^{x^2/2}\).
So
\[ y\,e^{x^2/2}=e^{x^2/2}+C \] Dividing through by \(e^{x^2/2}\):
\[ y=1+C\,e^{-x^2/2} \]
Step 4: Apply the initial condition to find C.
At \(x=0\), \(y=0\). Substituting:
\[ 0=1+C\,e^{0}=1+C \implies C=-1 \] So the particular solution is
\[ y=1-e^{-x^2/2} \]
Step 5: Evaluate at x = 1.0.
\[ y(1)=1-e^{-1/2}=1-e^{-0.5} \] Using \(e^{-0.5}\approx 0.6065\):
\[ y(1)\approx 1-0.6065=0.3935 \] Rounded to two decimal places, \(y(1)\approx 0.39\).

Final Answer:
The value of y at x = 1.0 is about 0.39. \[ \boxed{y(1)\approx 0.39} \]