Q21
jee_main_2024_01_february_morning
Linear Differential Equations
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
Numerical Answer. Answer: 14 to 14
Solution
### Related Formula
A first-order linear differential equation in standard form fracdxdt + P(t)x = Q(t) is solved using the Integrating Factor:
textI.F. = e^int P(t) \, dt
### Core Logic
Let's rearrange the given differential equation into standard linear form:
(t+1)dx = (2x + (t+1)^4)dt
fracdxdt = frac2x + (t+1)^4t+1
fracdxdt - frac2t+1x = (t+1)^3
### Step 1: Compute Integrating Factor and General Solution
Here, P(t) = -frac2t+1 and Q(t) = (t+1)^3.
textI.F. = e^int -frac2t+1 \, dt = e-2ln(t+1) = frac1(t+1)^2
The general solution is given by:
x cdot textI.F. = int Q(t) cdot textI.F. \, dt + C
fracx(t+1)^2 = int (t+1)^3 cdot frac1(t+1)^2 \, dt + C
fracx(t+1)^2 = int (t+1) \, dt + C = frac(t+1)^22 + C
### Step 2: Apply Boundary Condition
Given the initial condition x(0) = 2:
frac2(0+1)^2 = frac(0+1)^22 + C implies 2 = frac12 + C implies C = frac32
Hence, the specific solution curve is:
fracx(t+1)^2 = frac(t+1)^22 + frac32
x(t) = frac(t+1)^42 + frac32(t+1)^2
### Step 3: Evaluate at t = 1
Substituting t = 1:
x(1) = frac(1+1)^42 + frac32(1+1)^2 = frac162 + frac32(4) = 8 + 6 = 14
### Pattern Recognition
Sees: Linear form hidden under differential grouping coefficients.
Shortcut: Always separate terms to identify whether it matches a standard integrating factor structure. Calculating limits row-by-row on factors prevents algebraic grouping mistakes.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Mathematics: Differential Equations
