The differential equation is $\frac{dy}{dx} = 2x - y$ with the initial condition $y(1) = 0$. Let's solve this first-order linear differential equation.

Step 1: Rearrange the equation

We can rewrite the equation as: $\frac{dy}{dx} + y = 2x$

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

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by: $\mu(x) = e^{\int P(x)\, dx} = e^{\int 1\, dx} = e^x$

Step 3: Multiply the equation by the integrating factor

Now, multiply the entire differential equation by the integrating factor $e^x$: $e^x \frac{dy}{dx} + e^x y = 2xe^x$ This simplifies to: $\frac{d}{dx} \left( e^x y \right) = 2xe^x$

Step 4: Integrate both sides

Integrating both sides with respect to $x$: $e^x y = \int 2xe^x \, dx$

To solve $\int 2xe^x \, dx$, we use integration by parts, where $u = 2x$ and $dv = e^x dx$. Applying the formula $\int u \, dv = uv - \int v \, du$:

• $u = 2x$, so $du = 2 \, dx$
• $dv = e^x \, dx$, so $v = e^x$

Now, $\int 2xe^x \, dx = 2x e^x - \int 2e^x \, dx = 2x e^x - 2e^x$

Thus, the equation becomes: $e^x y = 2x e^x - 2e^x + C$

Step 5: Solve for $y$

Now, divide through by $e^x$: $y = 2x - 2 + Ce^{-x}$

Step 6: Apply the initial condition

We are given that $y(1) = 0$. Substituting $x = 1$ and $y = 0$ into the solution: $0 = 2(1) - 2 + Ce^{-1}$ $0 = 2 - 2 + Ce^{-1} = Ce^{-1}$ Thus, $C = 0$.

Final solution:

Therefore, the solution to the differential equation is: $y = 2x - 2$

Would you like further clarification or details on any step?

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Here are 5 related questions to expand your understanding:

1. How do you identify when a differential equation is linear?
2. What is an integrating factor, and why is it used in solving linear differential equations?
3. Can you solve $\frac{dy}{dx} = 3x - y$ with initial condition $y(0) = 2$?
4. How does the method of integration by parts work in more complex integrals?
5. What is the significance of the initial condition in solving differential equations?

Tip: When solving differential equations, always verify that your solution satisfies both the differential equation and the initial condition.