Q.Solution of $\frac{dy}{dx}-y=1$, $y(0)=1$ is given by:
(A) $xy=-e^x$
(B) $xy=-e^{-x}$
(C) $xy=-1$
(D) $y=2e^x-1$

This is an Initial Value Problem — a first-order linear ODE with a given condition.

Step 1: Solve the homogeneous part.

$\frac{dy}{dx} - y = 0$ gives $y_h = Ce^x$.

Step 2: Find a particular solution.

For the constant forcing term $1$, try $y_p = A$. Substituting: $0 - A = 1 \Rightarrow A = -1$.

So the general solution is $y = Ce^x - 1$. …

The problem gives us a first-order linear differential equation with an initial condition — an Initial Value Problem (IVP). The equation is $\frac{dy}{dx} - y = 1$, with $y(0) = 1$.

The key idea: when you see $\frac{dy}{dx} + P(x)y = Q(x)$, the standard tool is the integrating factor. Why does this work? Because multiplying both sides by a cleverly chosen function turns the left side into the derivative of a product, making it directly integrable. Here $P(x) = -1$ (constant), so the integrating factor is $e^{\int -1\,dx} = e^{-x}$.

Let’s work through it step by step.

1. Write the equation in standard form.

   It already is: $\frac{dy}{dx} - y = 1$. Here $P(x) = -1$, $Q(x) = 1$.

2. Compute the integrating factor.

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

3. Multiply the entire equation by $\mu(x)$.

Notice the left side is exactly $\frac{d}{dx}\left( y e^{-x} \right)$ — check by differentiating:

$\frac{d}{dx}(y e^{-x}) = \frac{dy}{dx}e^{-x} + y(-e^{-x}) = e^{-x}\frac{dy}{dx} - e^{-x}y$. Perfect.

4. Integrate both sides with respect to $x$.

5. Solve for $y$. Multiply through by $e^x$:

This is the general solution — a family of curves, one for each $C$.

6. Apply the initial condition $y(0) = 1$. Substitute $x = 0$, $y = 1$: …

Method: Solving an Initial Value Problem for a Linear Equation

An IVP couples a first-order linear equation with a condition. Solve the equation generally, then use the condition to pin the arbitrary constant.

Steps

Step 1: Write the general solution.

For $\frac{dy}{dx}-y=1$, the integrating factor is $e^{-x}$, giving $\frac{d}{dx}(ye^{-x})=e^{-x}$, so $y=Ce^{x}-1$.

Step 2: Apply the initial condition.

Substitute $y(0)=1$: $1=C-1\Rightarrow C=2$. …