If the solution of the differential equation dy/dx + eˣ(x² - 2)y = (x² - 2x)(x² - 2)e²ˣ satisfies y(

Q: Why can't we just separate the variables and integrate directly?

That is a great question! Separation of variables only works when you can isolate all $y$ terms on one side and all $x$ terms on the other. Here, the $y$ term is tied to a function of $x$ in a way that makes that impossible. This is a classic First-Order Linear Differential Equation, which requires the Integrating Factor method to solve.

Q: How do I know which part is $f(x)$ and which is $f'(x)$ in the integral?

It takes a bit of practice! Look for the term that, when differentiated, produces the other part. In our case, $x^2 - 2x$ is the obvious candidate for $f(x)$ because its derivative, $2x - 2$, is clearly related to the remaining terms. When you add $f(x)$ and $f'(x)$, you get $x^2 - 2$, which perfectly matches our expression.

Q: Do I really need to solve for $C$ before finding $y(2)$?

Absolutely! The general solution gives you a family of curves. The initial condition $y(0) = 0$ is the specific key that picks the one curve we care about. Without finding $C$, your solution is incomplete and you won't be able to determine the exact value of $y(2)$.

Question

If the solution of the differential equation $\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$ satisfies $y(0) = 0$, then the value of $y(2)$ is _______.

Options:

(A) -1

(B) 1

(C) 0

(D) e

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a differential equation that, at first glance, might seem like a tangled mess of exponentials and polynomials.

It is easy to feel overwhelmed when you see the following equation:

$$ \frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x} $$

But take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance.

Our first step is to recognize the structure. This is a classic **First-Order Linear Differential Equation**, which always follows the form:

$$ \frac{dy}{dx} + P(x)y = Q(x) $$

By identifying $P(x) = e^x(x^2 - 2)$ and $Q(x) = (x^2 - 2x)(x^2 - 2)e^{2x}$, we have already won the first battle. We have defined our battlefield.

### The Secret Weapon: The Integrating Factor

To solve this, we need our most reliable tool: the Integrating Factor ($I.F.$). The formula is defined as:

$$ I.F. = e^{\int P(x) dx} $$

Substituting our $P(x)$, we get $I.F. = e^{\int e^x(x^2 - 2) dx}$. Now, this integral looks daunting.

How do we integrate $e^x(x^2 - 2)$? This is where the magic happens. We use the beautiful property:

$$ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C $$

If we set $f(x) = x^2 - 2x$, then its derivative is $f'(x) = 2x - 2$. Adding them together, we get $(x^2 - 2x) + (2x - 2) = x^2 - 2$.

It fits perfectly! The integral evaluates to $e^x(x^2 - 2x)$. Thus, our Integrating Factor becomes:

$$ I.F. = e^{e^x(x^2 - 2x)} $$

It looks heavy, but it is the key to everything.

### The Path to Simplification

With our $I.F.$ in hand, the general solution is given by:

$$ y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C $$

Substituting our values, we get:

$$ y \cdot e^{e^x(x^2 - 2x)} = \int (x^2 - 2x)(x^2 - 2)e^{2x} \cdot e^{e^x(x^2 - 2x)} dx + C $$

This looks like a monster, but let's rearrange it. By splitting $e^{2x}$ into $e^x \cdot e^x$, we can group terms to make a substitution.

Let $t = e^x(x^2 - 2x)$. Differentiating this with respect to $x$ gives $dt = e^x(x^2 - 2) dx$.

Suddenly, the integral collapses into $\int t e^t dt$. This is a standard integral we can solve using Integration by Parts, resulting in $(t - 1)e^t + C$.

### The Final Victory

We are almost there. Substituting $t$ back, we get:

$$ y \cdot e^{e^x(x^2 - 2x)} = (e^x(x^2 - 2x) - 1) e^{e^x(x^2 - 2x)} + C $$

Using the initial condition $y(0) = 0$, we find $C = 1$.

Finally, to find $y(2)$, we plug in $x = 2$. Notice that $x^2 - 2x$ becomes $4 - 4 = 0$.

The entire expression simplifies beautifully to $y(2) \cdot 1 = -1 + 1$, which means **$y(2) = 0$**.

We have conquered the beast! Remember, the key is to stay calm, look for patterns, and trust the process. Keep practicing, and these structures will become second nature to you.

Animated Solution for Mathematics - Differential Equations: If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

Select Answer:

Visualized Solution

Identify the Type of Differential Equation

The Integrating Factor (I.F.) Formula

Solving the Integral $\int e^{x} (x^{2} - 2) d x$

Finalizing the Integrating Factor

The General Solution Setup

Simplifying the RHS Integral

Applying Substitution

Integration by Parts

The General Solution in terms of $x$

Finding the Constant $C$

Calculating $y (2)$

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Secret Weapon

The Path to Simplification

The Final Victory

Similar Questions

Let $y (x)$ be the solution of the differential equation $2 x^{2} d y + (e^{y} - 2 x) d x = 0, x > 0$ . If $y (e) = 1$ , then $y (1)$ is equal to :

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Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1$ , with $y (2) = 0$ . Then the value of $\frac{d y}{d x}$ at $x = 1$ is equal to:

Select Answer:

Visualized Solution

Identify the Type of Differential Equation

The Integrating Factor (I.F.) Formula

Solving the Integral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∫ex(x2−2)dx

Finalizing the Integrating Factor

The General Solution Setup

Simplifying the RHS Integral

Applying Substitution

Integration by Parts

The General Solution in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(2)

Conclusion and Key Takeaway

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Secret Weapon

The Path to Simplification

The Final Victory

Similar Questions

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