Solve the differential equation $\frac{dy}{dx}+1=e^{x+y}$.

Solve the differential equation (dy)/(dx)+1=ex+y.

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\frac{dy}{dx}+1=e^{x+y}

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Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

\frac{dy}{dx} = f(x) \cdot g(y)

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

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

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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The key idea is to use the substitution $u = x + y$ to turn the given equation into a separable form. The general solution is $e^{-(x+y)} + x = C$ .

Why this approach works

The equation $\frac{dy}{dx} + 1 = e^{x+y}$ is not directly separable — the variables $x$ and $y$ are tangled inside the exponential. But notice the combination $x+y$ appears. Whenever you see an expression like $f(x+y)$ or $f(ax+by)$ , a substitution $u = x+y$ (or $u = ax+by$ ) often simplifies the derivative as well, because $\frac{du}{dx} = 1 + \frac{dy}{dx}$ . That turns the left-hand side into something neat.

This is a classic Initial Value Problem technique: transform a non-separable equation into a separable one by a clever change of variable.

Step-by-step solution

Set up the substitution Let $u = x + y$ . Then differentiate with respect to $x$ :

\frac{du}{dx} = 1 + \frac{dy}{dx}.

The given equation is $\frac{dy}{dx} + 1 = e^{x+y}$ . The left side is exactly $\frac{du}{dx}$ , and the right side becomes $e^u$ . So the equation simplifies to:

\frac{du}{dx} = e^u.

Separate variables This is now a separable differential equation:

\frac{du}{e^u} = dx \quad \Rightarrow \quad e^{-u} \, du = dx.

Integrate both sides

\int e^{-u} \, du = \int dx.

The left integral gives $-e^{-u}$ , and the right gives $x + C$ :

-e^{-u} = x + C.

Back-substitute $u = x + y$

-e^{-(x+y)} = x + C.

Multiply both sides by $-1$ to get a cleaner form:

$e^{-(x+y)} = -x - C.$ …

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Q.Solve the differential equation $\frac{dy}{dx}+1=e^{x+y}$ .

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Why this approach works

Step-by-step solution

Method: Reducing to separable by the substitution $x+y=t$

Steps

Common Mistakes

Q.Solve the differential equation dydx+1=ex+y\frac{dy}{dx}+1=e^{x+y}dxdy​+1=ex+y.

Why this approach works

Step-by-step solution

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Q.Solve the differential equation $\frac{dy}{dx}+1=e^{x+y}$ .

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