Solving Differential Equation: $\frac{dy}{dx} + xy = x$

The given differential equation is a first-order linear equation:

$ \frac{dy}{dx} + xy = x $

This equation is in the standard form $ \frac{dy}{dx} + P(x)y = Q(x) $. Here:

• $ P(x) = x $
• $ Q(x) = x $

Integrating Factor Calculation

The integrating factor (IF) is calculated using the formula:

$ IF = e^{\int P(x) dx} $

Substitute $ P(x) = x $:

$ IF = e^{\int x dx} = e^{\frac{x^2}{2}} $

General Solution Derivation

Multiply the differential equation by the integrating factor $ e^{\frac{x^2}{2}} $:

$ e^{\frac{x^2}{2}} \frac{dy}{dx} + x e^{\frac{x^2}{2}} y = x e^{\frac{x^2}{2}} $

The left side can be written as the derivative of $ y \cdot IF $:

$ \frac{d}{dx} \left( y \cdot e^{\frac{x^2}{2}} \right) = x e^{\frac{x^2}{2}} $

Integrate both sides with respect to $ x $:

$ y \cdot e^{\frac{x^2}{2}} = \int x e^{\frac{x^2}{2}} dx $

Use substitution $ u = \frac{x^2}{2} $, $ du = x dx $ for the integral:

$ \int e^u du = e^u + C = e^{\frac{x^2}{2}} + C $

So, the equation becomes:

$ y \cdot e^{\frac{x^2}{2}} = e^{\frac{x^2}{2}} + C $

The general solution is obtained by dividing by $ e^{\frac{x^2}{2}} $:

$ y = 1 + C e^{-\frac{x^2}{2}} $

Applying Initial Condition

Use the given condition $ y = 0 $ when $ x = 0 $ to find the constant $ C $:

$ 0 = 1 + C e^{-\frac{0^2}{2}} $

$ 0 = 1 + C \cdot 1 $

$ C = -1 $

Specific Solution Determination

Substitute $ C = -1 $ back into the general solution:

$ y = 1 - e^{-\frac{x^2}{2}} $

Value Calculation at $x = 1.0$

Calculate the value of $ y $ when $ x = 1.0 $:

$ y(1.0) = 1 - e^{-\frac{(1.0)^2}{2}} = 1 - e^{-\frac{1}{2}} = 1 - e^{-0.5} $

Calculate the numerical value:

$ y(1.0) \approx 1 - 0.60653 = 0.39347 $

Final Answer Rounding

Round the result to two decimal places as required:

$ y(1.0) \approx 0.39 $