Given that $\frac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5$. Find the value of $x$ when $y=3$.

> [!TLDR] > This is a separable first-order ODE. Separate variables, integrate both sides, use the initial condition to find the constant, then solve for $x$ when $y=3$. The result is $x = \frac{e^6 + 9}{2}$. The key idea here is **Separation of Variables**. When a differential equation can be written so that all terms involving $y$ (including $dy$) are on one side and all terms involving $x$ (including $dx$) are on the other, you can integrate each side independently. That’s exactly what we have here: $\frac{dy}{dx} = e^{-2y}$ is a product of a function of $y$ and a constant function of $x$ (in fact, just $1$). So we can “separate” by multiplying both sides by $dx$ and by $e^{2y}$. Let’s walk through it. 1. **Separate the variables.** Multiply both sides by $dx$ and by $e^{2y}$: $$ e^{2y} \, dy = dx $$ Now the left side depends only on $y$, the right side only on $x$. 2. **Integrate both sides.** $$ \int e^{2y} \, dy = \int dx $$ The left integral is $\frac{1}{2} e^{2y} + C_1$ (since the derivative of $e^{2y}$ is $2e^{2y}$). The right integral is $x + C_2$. Combine the constants into a single constant $C$: $$ \frac{1}{2} e^{2y} = x + C $$ 3. **Apply the initial condition $y=0$ when $x=5$.** Substitute: $$ \frac{1}{2} e^{0} = 5 + C \quad \Rightarrow \quad \frac{1}{2} = 5 + C $$ So $C = \frac{1}{2} - 5 = -\frac{9}{2}$. 4. **Write the particular solution.** $$ \frac{1}{2} e^{2y} = x - \frac{9}{2} $$ Multiply through by 2 to simplify: $$ e^{2y} = 2x - 9 $$ 5. **Find $x$ when $y=3$.** Substitute $y=3$: $$ e^{6} = 2x - 9 $$ Solve for $x$: $$ 2x = e^{6} + 9 \quad \Rightarrow \quad x = \frac{e^{6} + 9}{2} $$ > [!WARNING] > A common mistake is forgetting to handle the constant of integration correctly. If you combine constants into one, always check that your final equation satisfies the initial condition. Also, note that $e^{-2y}$ is never zero, so no singular

Given that (dy)/(dx)=e-2y and y=0 when x=5. Find the value of x when y=3.

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\frac{dy}{dx}=e^{-2y}

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

Tip

Always check if $g(y) = 0$ gives a solution. Here $g(y) = y$ , so $y = 0$ is also a solution (the trivial one). Our initial condition $y(0)=3$ picks the non-zero branch.

When Does It Apply?

Separation of Variables works for first-order ODEs of the form $\frac{dy}{dx} = f(x)g(y)$ , and for certain partial differential equations like the heat equation (a more advanced use — the idea is the same: assume the solution is a product of functions of each variable). It does not work for equations where $x$ and $y$ are added, subtracted, or composed in non-product ways, or for higher-order ODEs (generally).

The Big Picture

Separation of Variables is your first real tool for solving differential equations. It reduces a problem with two moving parts into two independent single-variable integrals: you separate the variables, handle each alone, then reassemble with the initial condition. Think of it as untangling a knot by pulling the two ends apart — once separate, each piece is easy to deal with.

Separation of Variables is the very first solving technique taught in NCERT's Class 12 Differential Equations chapter and one of the most heavily tested skills in CBSE board exams and JEE Main. Students searching "separable differential equations important questions" or "differential equations class 12 formula" will find this method the starting point for almost every other technique in the chapter.

This is a separable first-order ODE. Separate variables, integrate both sides, use the initial condition to find the constant, then solve for $x$ when $y=3$ . The result is $x = \frac{e^6 + 9}{2}$ .

The key idea here is Separation of Variables. When a differential equation can be written so that all terms involving $y$ (including $dy$ ) are on one side and all terms involving $x$ (including $dx$ ) are on the other, you can integrate each side independently. That’s exactly what we have here: $\frac{dy}{dx} = e^{-2y}$ is a product of a function of $y$ and a constant function of $x$ (in fact, just $1$ ). So we can “separate” by multiplying both sides by $dx$ and by $e^{2y}$ .

Let’s walk through it.

Separate the variables. Multiply both sides by $dx$ and by $e^{2y}$ :

e^{2y} \, dy = dx

Now the left side depends only on $y$ , the right side only on $x$ .

Integrate both sides.

\int e^{2y} \, dy = \int dx

The left integral is $\frac{1}{2} e^{2y} + C_1$ (since the derivative of $e^{2y}$ is $2e^{2y}$ ). The right integral is $x + C_2$ . Combine the constants into a single constant $C$ :

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

Apply the initial condition $y=0$ when $x=5$ . Substitute:

\frac{1}{2} e^{0} = 5 + C \quad \Rightarrow \quad \frac{1}{2} = 5 + C

So $C = \frac{1}{2} - 5 = -\frac{9}{2}$ .

Write the particular solution.

\frac{1}{2} e^{2y} = x - \frac{9}{2}

Multiply through by 2 to simplify:

e^{2y} = 2x - 9

Find $x$ when $y=3$ . Substitute $y=3$ :

e^{6} = 2x - 9

Solve for $x$ :

2x = e^{6} + 9 \quad \Rightarrow \quad x = \frac{e^{6} + 9}{2}

Watch out

A common mistake is forgetting to handle the constant of integration correctly. If you combine constants into one, always check that your final equation satisfies the initial condition. Also, note that $e^{-2y}$ is never zero, so no singular solutions are lost here.

Tip

You can check your answer by differentiating implicitly: from $e^{2y} = 2x - 9$ , differentiate both sides with respect to $x$ to get $2e^{2y} \frac{dy}{dx} = 2$ , which simplifies to $\frac{dy}{dx} = e^{-2y}$ . Works perfectly.

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Q.Given that $\frac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5$ . Find the value of $x$ when $y=3$ .

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

When Does It Apply?

The Big Picture

Method: Solving a separable initial-value problem, then evaluating

Steps

Common Mistakes

Q.Given that dydx=e−2y\frac{dy}{dx}=e^{-2y}dxdy​=e−2y and y=0y=0y=0 when x=5x=5x=5. Find the value of xxx when y=3y=3y=3.

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Q.Given that $\frac{dy}{dx}=e^{-2y}$ and $y=0$ when $x=5$ . Find the value of $x$ when $y=3$ .

Unlock everything free for 14 days