Let y'(x) + y(x) g'(x) = g(x) g'(x), y(0) = 0, x ∈ ℝ, where f'(x) denotes df(x)/dx and g(x) is a giv

Q: Wait, why can't we just separate the variables? It looks like it might work.

Great intuition, but look closely! The term $y(x)g'(x)$ couples $y$ and $x$ in a way that prevents simple separation. The structure $y' + P(x)y = Q(x)$ is the hallmark of a linear differential equation, which requires the integrating factor method to solve.

Q: Do I need to know what $g(x)$ is to solve this?

Not at all! That is the beauty of this problem. The problem is designed to test your ability to manipulate the structure of the equation rather than solving for a specific function. As long as you know the values at the boundaries, the abstract nature of $g(x)$ does not stop us.

Q: Why did we get $C=1$?

We found $C$ by applying the initial condition $y(0)=0$. By substituting $x=0$ into our general solution and using the given $g(0)=0$, the equation simplifies to $0 = (0-1)e^0 + C$, which forces $C=1$. Always remember to use your initial conditions to lock down that constant!

Question

Let $y'(x) + y(x) g'(x) = g(x) g'(x)$, $y(0) = 0$, $x \in \mathbb{R}$, where $f'(x)$ denotes $\frac{df(x)}{dx}$ and $g(x)$ is a given non-constant differentiable function on $\mathbb{R}$ with $g(0) = g(2) = 0$. Then the value of $y(2)$ is

Sigma Percentile · Accepted Answer

### The Beauty of Abstract Functions

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that looks intimidating but is actually a masterpiece of elegance.

We are faced with the differential equation:

$$y'(x) + y(x) g'(x) = g(x) g'(x)$$

At first glance, the presence of an unknown, non-constant function $g(x)$ might make you want to reach for a calculator or give up. But hold on! In the world of JEE Advanced, we don't need to know the identity of $g(x)$ to understand its behavior. We just need to respect its structure.

### Phase 1: The Recognition

Let's look at the equation again: $y'(x) + g'(x)y(x) = g(x)g'(x)$. This is the classic form of a **First Order Linear Differential Equation**:

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

By comparing our equation to this standard form, we can immediately identify our components: $P(x) = g'(x)$ and $Q(x) = g(x)g'(x)$. The fog is already lifting, isn't it?

### Phase 2: The Integrating Factor

Now, we need our secret weapon: the Integrating Factor ($IF$). The formula is:

$$IF = e^{\int P(x) dx}$$

Substituting our $P(x)$, we get $IF = e^{\int g'(x) dx}$. This is where the magic happens.

The integral of a derivative $g'(x)$ is simply the function $g(x)$ itself. Thus, our integrating factor simplifies beautifully to:

$$IF = e^{g(x)}$$

This factor is the key that unlocks the entire equation.

### Phase 3: The Integration

With our $IF$ in hand, we write the general solution:

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

Substituting our $Q(x)$, we get:

$$y e^{g(x)} = \int g(x) e^{g(x)} g'(x) dx + C$$

This integral looks daunting, but let's use the substitution method. Let $g(x) = t$, which implies $g'(x) dx = dt$.

The integral transforms into the much friendlier $\int t e^t dt$. Using integration by parts, where $u=t$ and $dv=e^t dt$, we get:

$$\int t e^t dt = t e^t - \int e^t dt = (t-1)e^t$$

Substituting back $t = g(x)$, our general solution becomes:

$$y e^{g(x)} = (g(x)-1)e^{g(x)} + C$$

### Phase 4: The Final Reveal

We are almost there! We are given $y(0) = 0$ and $g(0) = 0$. Plugging these into our general solution:

$$0 \cdot e^0 = (0-1)e^0 + C \Rightarrow 0 = -1 + C \Rightarrow C = 1$$

Our particular solution is:

$$y(x) = g(x) - 1 + e^{-g(x)}$$

Finally, we need $y(2)$. Given $g(2) = 0$, we substitute $x=2$:

$$y(2) = g(2) - 1 + e^{-g(2)} = 0 - 1 + e^0 = 0$$

And there it is—the **answer is 0**. It is a beautiful, clean result that proves that even in the face of abstract functions, logic and structure will always guide you to the truth.

Enter Numerical Value:

Visualized Solution

Identifying the Linear Differential Equation

Calculating the Integrating Factor $e^{g (x)}$

Setting up the General Solution

Solving the Integral using Substitution $g (x) = t$

Applying Integration by Parts

Determining the Constant $C$ using $y (0) = 0$

The Particular Solution Curve

Evaluating $y (2)$ using $g (2) = 0$

Final Result and Key Takeaways

The Sigma Insight: Linear Differential Equations

The Beauty of Abstract Functions

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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

If $y = y (x)$ is the solution of the equation $e^{s i n y} cos y \frac{d y}{d x} + e^{s i n y} cos x = cos x, y (0) = 0$ ; then $1 + y (\frac{π}{2}) + \frac{π}{2} y^{'} (\frac{π}{2}) + \frac{1}{2} y^{''} (\frac{π}{2})$ is equal to

Let $y = y (x)$ be a differentiable function in the interval $(0, \infty)$ such that $y (1) = 2$ , and $lim_{t \to x} \frac{t ^{2} y ( x ) - x ^{2} y ( t )}{x - t} = 3$ for each $x > 0$ . Then $2 y (2)$ is equal to

Let $y = f (x)$ is the solution of the differential equation $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to:

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

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:

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y = f (x)$ , where $f (x) = {1, 0, x \in [0, 1] otherwise$ . If $y (0) = 0$ , then $y (3/2)$ is :

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Identifying the Linear Differential Equation

Calculating the Integrating Factor .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } eg(x)

Setting up the General Solution

Solving the Integral using Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)=t

Applying Integration by Parts

Determining the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(0)=0

The Particular Solution Curve

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

Final Result and Key Takeaways

The Sigma Insight: Linear Differential Equations

The Beauty of Abstract Functions

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(e) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the equation esinycosydxdy​+esinycosx=cosx,y(0)=0; then 1+y(2π​)+2π​​y′(2π​)+2​1​y′′(2π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be a differentiable function in the interval (0,∞) such that y(1)=2, and limt→x​x−tt2y(x)−x2y(t)​=3 for each x>0. Then 2y(2) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=f(x) is the solution of the differential equation ey(dxdy​−1)=ex such that y(0)=0, then y(1) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation, dxdy​+ytanx=2x+x2tanx, x∈(−π/2,π/2), such that y(0)=1. Then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​=(y+1)((y+1)ex2/2−x),0<x<2.1, with y(2)=0. Then the value of dxdy​ at x=1 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0. If y(0)=0, then y(2) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​+2y=f(x), where f(x)={1,0,​x∈[0,1]otherwise​. If y(0)=0, then y(3/2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation 2 cos xdxdy​=sin 2x−4y sin x,x∈(0,2π​). If y(3π​)=0 then y′(4π​)+y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function such that f′(x)+f(x)=∫02​f(t)dt. If f(0)=e−2, then 2f(0)−f(2) is equal to ______.