If x = x(t) is the solution of the differential equation (t + 1)dx = (2x + (t + 1)⁴) dt, x(0) = 2, t

Q: Why is the equation considered linear? I see a $(t+1)^4$ term, which looks non-linear.

Great observation! A differential equation is linear in $x$ if $x$ and its derivative $\frac{dx}{dt}$ appear only to the first power and are not multiplied together. The term $(t+1)^4$ is a function of $t$ only, so it acts as our $Q(t)$. It doesn't affect the linearity with respect to $x$! Think of it as a forcing function that depends only on time.

Q: I keep getting the wrong Integrating Factor. Why is the negative sign so important?

The negative sign is the heartbeat of this problem. If you miss it, your IF becomes $(t+1)^2$ instead of $\frac{1}{(t+1)^2}$. This changes the entire integration process. Always carry the sign with the coefficient of $x$ when identifying $P(t)$. In this case, $P(t) = -\frac{2}{t+1}$, and that negative sign is what allows the terms to cancel out perfectly later.

Q: Do I really need to find $C$ before calculating $x(1)$?

Absolutely. The general solution gives you a family of curves. The initial condition $x(0)=2$ acts as a 'pin' that selects the one specific curve passing through that point. Without finding $C$, you are just describing a shape, not the specific path. You must anchor your solution to the given coordinate before you can predict the future value at $t=1$.

Question

If $x = x(t)$ is the solution of the differential equation $(t + 1)dx = (2x + (t + 1)^4) dt, x(0) = 2$, then, $x(1)$ equals

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE Advanced path. Today, we are going to demystify a problem that often strikes fear into the hearts of students: the linear differential equation.

The given equation is $(t + 1)dx = (2x + (t + 1)^4) dt$. While it may initially appear chaotic, mathematics is about recognizing patterns.

### The Art of Rearrangement

To solve this, we must transform the equation into the standard form of a linear differential equation: 
$$\frac{dx}{dt} + P(t)x = Q(t)$$

We begin by dividing the entire equation by $(t+1)dt$:
$$\frac{dx}{dt} = \frac{2x}{(t+1)} + (t+1)^3$$

Rearranging the terms to isolate the $x$ component, we obtain:
$$\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$$

Here, we identify our functions as $P(t) = -\frac{2}{t+1}$ and $Q(t) = (t+1)^3$.

### The Magic of the Integrating Factor

The Integrating Factor ($IF$) is the key to unlocking this equation. It is defined by the formula:
$$IF = e^{\int P(t) dt}$$

Substituting our $P(t)$, we calculate:
$$IF = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln(t+1)}$$

Using the properties of logarithms, we simplify the exponent:
$$IF = e^{\ln(t+1)^{-2}} = (t+1)^{-2} = \frac{1}{(t+1)^2}$$

### The Symphony of Cancellation

We now apply the general solution formula: $x \cdot (IF) = \int Q(t) \cdot (IF) dt$. Substituting our known values:
$$x \cdot \frac{1}{(t+1)^2} = \int (t+1)^3 \cdot \frac{1}{(t+1)^2} dt$$

The right side simplifies beautifully:
$$\frac{x}{(t+1)^2} = \int (t+1) dt$$

Integrating with respect to $t$, we find:
$$\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + C$$

### Finding the Anchor

We use the initial condition $x(0) = 2$ to determine the constant $C$. Substituting these values into our general solution:
$$\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C$$

This simplifies to $2 = \frac{1}{2} + C$, which yields $C = \frac{3}{2}$. Thus, the specific solution is:
$$\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + \frac{3}{2}$$

Multiplying by $(t+1)^2$, we get the function:
$$x(t) = \frac{(t+1)^4}{2} + \frac{3(t+1)^2}{2}$$

### The Final Victory

To find $x(1)$, we substitute $t=1$ into our specific solution:
$$x(1) = \frac{(1+1)^4}{2} + \frac{3(1+1)^2}{2}$$

$$x(1) = \frac{2^4}{2} + \frac{3 \cdot 2^2}{2} = \frac{16}{2} + \frac{12}{2}$$

$$x(1) = 8 + 6 = 14$$

The final answer is **14**. You have successfully navigated the path from confusion to clarity.

Animated Solution for Mathematics - Differential Equations: If $x = x (t)$ is the solution of the differential equation $(t + 1) d x = (2 x + (t + 1)^{4}) d t, x (0) = 2$ , then, $x (1)$ equals

Enter Numerical Value:

Visualized Solution

Identify the Differential Equation

Rearranging to Standard Form

Identify $P (t)$ and $Q (t)$

The Integrating Factor (IF) Setup

Evaluating the IF

General Solution Setup

Simplifying the Integral

Initial Condition: Finding C

Solving for C

The Specific Solution

Final Calculation: $x (1)$

The Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Rearrangement

The Magic of the Integrating Factor

The Symphony of Cancellation

Finding the Anchor

The Final Victory

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt,x(0)=2, then, x(1) equals