general solution of the differential equation, $$ d y / d x+x y=x $$, is

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To solve the differential equation:

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

we recognize it as a first-order linear ordinary differential equation of the form:

$$
\frac{dy}{dx} + P(x) y = Q(x)
$$

where $$ P(x) = x $$ and $$ Q(x) = x $$.

### Step 1: Determine the Integrating Factor
The integrating factor $$ \mu(x) $$ is given by:

$$
\mu(x) = e^{\int P(x) \, dx} = e^{\int x \, dx} = e^{\frac{x^2}{2}}
$$

### Step 2: Multiply Through by the Integrating Factor
Multiply every term in the differential equation by $$ \mu(x) $$:

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

This simplifies to:

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

### Step 3: Integrate Both Sides
Integrate both sides with respect to $$ x $$:

$$
y e^{\frac{x^2}{2}} = \int x e^{\frac{x^2}{2}} \, dx + C
$$

Let $$ u = \frac{x^2}{2} $$, hence $$ du = x \, dx $$. The integral becomes:

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

So,

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

### Step 4: Solve for $$ y $$
Divide both sides by $$ e^{\frac{x^2}{2}} $$:

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

### Final Answer

After simplifying, all solutions are 1 plus a constant times e^–½x². Thus,

y = 1 + C e^−x²⁄2
The general solution to the differential equation $$ \frac{dy}{dx} + x y = x $$ is: $$ y = 1 + C e^{-\frac{x^2}{2}} $$ where $$ C $$ is an arbitrary constant.

general solution of the differential equation, , is

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