Can you do the same problem but dy/dx = 2x-y when y=0 and x= 1

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.

Can you do the same problem but dy/dx = 2x - y when y = 0 and x = 1?

Explore how to solve the first-order linear differential equation dy/dx = 2x - y with the initial condition y(1) = 0. This solution includes the use of integrating factors and integration by parts, explained step by step. Ideal for undergraduate students studying differential equations.

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