Solution & Explanation
### Related Formula
For a linear differential equation fracdydx + Py = Q$\frac{dy}{dx} + Py = Q$, the Integrating Factor (IF) is defined as:
textIF = e^int P \, dx$$\text{IF} = e^{\int P \, dx}$$
### Core Logic
Divide the full differential equation by (x^2+1)$(x^2+1)$:
fracdydx - left(frac2xx^2+1right)y = frac(x^2+1)^2 cos xx^2+1 = (x^2+1)cos x$$\frac{dy}{dx} - \left(\frac{2x}{x^2+1}\right)y = \frac{(x^2+1)^2 \cos x}{x^2+1} = (x^2+1)\cos x$$
This is a standard Linear Differential Equation with:
P = -frac2xx^2+1, quad Q = (x^2+1)cos x$$P = -\frac{2x}{x^2+1}, \quad Q = (x^2+1)\cos x$$
textIF = e^int -frac2xx^2+1\,dx = e^-ln(x^2+1) = frac1x^2+1$$\text{IF} = e^{\int -\frac{2x}{x^2+1}\,dx} = e^{-\ln(x^2+1)} = \frac{1}{x^2+1}$$
### Step 1: Solve for General Solution
The solution format is y cdot textIF = int Q cdot textIF \, dx$y \cdot \text{IF} = \int Q \cdot \text{IF} \, dx$:
y cdot frac1x^2+1 = int (x^2+1)cos x cdot frac1x^2+1 \, dx$$y \cdot \frac{1}{x^2+1} = \int (x^2+1)\cos x \cdot \frac{1}{x^2+1} \, dx$$
fracyx^2+1 = sin x + c$$\frac{y}{x^2+1} = \sin x + c$$
Using the boundary condition y(0) = 1$y(0) = 1$:
frac10+1 = sin(0) + c implies c = 1$$\frac{1}{0+1} = \sin(0) + c \implies c = 1$$
y = (x^2+1)(sin x + 1)$$y = (x^2+1)(sin x + 1)$$
### Step 2: Definite Integration Evaluation
We need to evaluate int_-3^3 y \, dx$\int_{-3}^{3} y \, dx$:
int_-3^3 (x^2+1)(sin x + 1) \, dx = int_-3^3 (x^2sin x + x^2 + sin x + 1) \, dx$$\int_{-3}^{3} (x^2+1)(sin x + 1) \, dx = \int_{-3}^{3} (x^2\sin x + x^2 + \sin x + 1) \, dx$$
By symmetry of odd/even functions over symmetric intervals [-a, a]$[-a, a]$:
int_-3^3 x^2sin x \, dx = 0$\int_{-3}^{3} x^2\sin x \, dx = 0$ (since it is an odd function)
int_-3^3 sin x \, dx = 0$\int_{-3}^{3} \sin x \, dx = 0$ (since it is an odd function)
Thus, we are left with the even components:
int_-3^3 (x^2 + 1) \, dx = 2 int_0^3 (x^2 + 1) \, dx = 2 left[ fracx^33 + x right]_0^3 = 2(9 + 3) = 24$$\int_{-3}^{3} (x^2 + 1) \, dx = 2 \int_{0}^{3} (x^2 + 1) \, dx = 2 \left[ \frac{x^3}{3} + x \right]_{0}^{3} = 2(9 + 3) = 24$$
### Pattern Recognition
Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Differential Equations
Class 12 Mathematics: Integral Calculus
More Differential Equations Previous-Year Questions — Page 3
Q62
jee_main_2025_03_april_morning
Linear Differential Equations
Let g$g$ be a differentiable function such that int_0^xg(t)dt=x-int_0^xtg(t)dt$\int_{0}^{x}g(t)dt=x-\int_{0}^{x}tg(t)dt$ [cite: 568], xge0$x\ge0$ [cite: 569] and let y=y(x)$y=y(x)$ satisfy the differential equation fracdydx - ytan x = 2(x+1)sec x \, g(x)$\frac{dy}{dx} - y\tan x = 2(x+1)\sec x \, g(x)$ [cite: 571, 575, 578, 581], xin[0,fracpi2)$x\in[0,\frac{\pi}{2})$[cite: 581]. If y(0)=0$y(0)=0$ [cite: 579] then yleft(fracpi3right)$y\left(\frac{\pi}{3}\right)$ is equal to[cite: 582]:
- A. frac2pi3sqrt3$\frac{2\pi}{3\sqrt{3}}$
- B. frac4pi3$\frac{4\pi}{3}$
- C. frac2pi3$\frac{2\pi}{3}$
- D. frac4pi3sqrt3$\frac{4\pi}{3\sqrt{3}}$
Solution
### Related Formula
Leibniz Integral Rule for differentiation:
fracmathrmdmathrmdxleft(int_0^x f(t)mathrmdtright) = f(x)$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^x f(t)\mathrm{d}t\right) = f(x)$$
### Core Logic
Differentiate the given integral relation using Leibniz rule [cite: 1344]:
fracmathrmdmathrmdxleft[int_0^xg(t)dtright] = fracmathrmdmathrmdxleft[x-int_0^xtg(t)dtright]$$\frac{\mathrm{d}}{\mathrm{d}x}\left[\int_{0}^{x}g(t)dt\right] = \frac{\mathrm{d}}{\mathrm{d}x}\left[x-\int_{0}^{x}tg(t)dt\right]$$ [cite: 1344]
g(x) = 1 - xg(x) implies g(x)(1+x) = 1 implies g(x) = frac11+x$$g(x) = 1 - xg(x) \implies g(x)(1+x) = 1 \implies g(x) = \frac{1}{1+x}$$ [cite: 1345]
Substitute g(x)$g(x)$ into target differential equation configuration [cite: 1346]:
fracmathrmdymathrmdx - ytan x = 2(x+1)sec x cdot left(frac11+xright) = 2sec x$$\frac{\mathrm{d}y}{\mathrm{d}x} - y\tan x = 2(x+1)\sec x \cdot \left(\frac{1}{1+x}\right) = 2\sec x$$ [cite: 1346]
### Step 1: Finding the Integrating Factor
This matches a linear form fracmathrmdymathrmdx + P(x)y = Q(x)$\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)$ where P(x) = -tan x$P(x) = -\tan x$.
textI.F. = e^int -tan x \, mathrmdx = e^ln|cos x| = cos x$$\text{I.F.} = e^{\int -\tan x \, \mathrm{d}x} = e^{\ln|\cos x|} = \cos x$$ [cite: 1346]
Write general functional solution template [cite: 1348]:
y cdot cos x = int (2sec x cdot cos x) \, mathrmdx = int 2 \, mathrmdx = 2x + C$$y \cdot \cos x = \int (2\sec x \cdot \cos x) \, \mathrm{d}x = \int 2 \, \mathrm{d}x = 2x + C$$ [cite: 1348]
Given boundary condition y(0) = 0 implies 0 = 0 + C implies C = 0$y(0) = 0 \implies 0 = 0 + C \implies C = 0$ [cite: 1349].
y(x) = frac2xcos x = 2xsec x$$y(x) = \frac{2x}{\cos x} = 2x\sec x$$ [cite: 1350]
### Step 2: Numeric substitution
Substitute variable parameter values x = fracpi3$x = \frac{\pi}{3}$ [cite: 1352]:
yleft(fracpi3right) = 2left(fracpi3right)secleft(fracpi3right) = frac2pi3 cdot 2 = frac4pi3$$y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \cdot 2 = \frac{4\pi}{3}$$ [cite: 1351]
### Pattern Recognition
Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Differential Equations
Q66
jee_main_2025_04_april_evening
Linear Differential Equations
If a curve y = y(x)$y = y(x)$ passes through the point left(1, fracpi2right)$\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation (7x^4 cot y - e^x mathrmcosec\,y) fracdxdy = x^5$(7x^4 \cot y - e^x \mathrm{cosec}\,y) \frac{dx}{dy} = x^5$, x geq 1$x \geq 1$, then at x = 2$x = 2$, the value of cosine is:
- A. frac2mathrme^2 - mathrme64$\frac{2\mathrm{e}^2 - \mathrm{e}}{64}$
- B. frac2mathrme^2 + mathrme64$\frac{2\mathrm{e}^2 + \mathrm{e}}{64}$
- C. frac2mathrme^2 - mathrme128$\frac{2\mathrm{e}^2 - \mathrm{e}}{128}$
- D. frac2mathrme^2 + mathrme128$\frac{2\mathrm{e}^2 + \mathrm{e}}{128}$
Solution
### Core Logic
Let's rearrange the given differential equation by expressing fracdydx$\frac{dy}{dx}$:
x^5 fracdydx = 7x^4 cot y - e^x csc y$$x^5 \frac{dy}{dx} = 7x^4 \cot y - e^x \csc y$$
Dividing both sides by x^5$x^5$:
fracdydx = frac7x cot y - frace^xx^5 csc y$$\frac{dy}{dx} = \frac{7}{x} \cot y - \frac{e^x}{x^5} \csc y$$
Multiply the entire equation by sin y$\sin y$ to clear denominators:
sin y fracdydx - frac7x cos y = -frace^xx^5$$\sin y \frac{dy}{dx} - \frac{7}{x} \cos y = -\frac{e^x}{x^5}$$
This can be transformed into a linear form by substituting t = -cos y$t = -\cos y$. Then fracdtdx = sin y fracdydx$\frac{dt}{dx} = \sin y \frac{dy}{dx}$.
### Step 1: Solving the Linear ODE
Substituting t$t$ leads to:
fracdtdx + frac7x t = -frace^xx^5$$\frac{dt}{dx} + \frac{7}{x} t = -\frac{e^x}{x^5}$$
This is a standard linear first-order ODE with P(x) = frac7x$P(x) = \frac{7}{x}$. The Integrating Factor (I.F.) is:
textI.F. = e^int frac7x dx = e^7 ln x = x^7$$\text{I.F.} = e^{\int \frac{7}{x} dx} = e^{7 \ln x} = x^7$$
The general solution is:
t cdot x^7 = int left(-frace^xx^5right) cdot x^7 dx = -int x^2 e^x dx$$t \cdot x^7 = \int \left(-\frac{e^x}{x^5}\right) \cdot x^7 dx = -\int x^2 e^x dx$$
### Step 2: Evaluating the Integration and Constant
Using integration by parts for int x^2 e^x dx$\int x^2 e^x dx$:
int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x$$\int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x$$
Substituting this back:
-cos y cdot x^7 = -e^x(x^2 - 2x + 2) + C$$-\cos y \cdot x^7 = -e^x(x^2 - 2x + 2) + C$$
cos y cdot x^7 = e^x(x^2 - 2x + 2) - C$$\cos y \cdot x^7 = e^x(x^2 - 2x + 2) - C$$
Since the curve passes through left(1, fracpi2right)$\left(1, \frac{\pi}{2}\right)$:
cosleft(fracpi2right) cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C implies 0 = e(1) - C implies C = e$$\cos\left(\frac{\pi}{2}\right) \cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C \implies 0 = e(1) - C \implies C = e$$
### Step 3: Calculating cos y at x = 2
Now substitute x = 2$x = 2$ and C = e$C = e$ into our equation block:
cos y cdot (2^7) = e^2(2^2 - 2(2) + 2) - e$$\cos y \cdot (2^7) = e^2(2^2 - 2(2) + 2) - e$$
cos y cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e$$\cos y \cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e$$
cos y = frac2e^2 - e128$$\cos y = \frac{2e^2 - e}{128}$$
### Pattern Recognition
When trigonometric terms are mixed inside an ODE containing derivative blocks like fracdxdy$\frac{dx}{dy}$ or fracdydx$\frac{dy}{dx}$, check if clearing denominators using sin y$\sin y$ or cos y$\cos y$ reveals a standard substitution path for a Linear ODE.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Differential Equations
Q58
jee_main_2025_04_april_morning
Area Bounded by Curves
Let f colon [0, infty) to mathbbR$f \colon [0, \infty) \to \mathbb{R}$ be a differentiable function such that f(x) = 1 - 2x + int_0^x e^x - t f(t) mathrmdt$f(x) = 1 - 2x + \int_0^x e^{x - t} f(t) \mathrm{d}t$ for all x in [0, infty)$x \in [0, \infty)$. Then the area of the region bounded by y = f(x)$y = f(x)$ and the coordinate axes is
- A. sqrt5$\sqrt{5}$
- B. frac12$\frac{1}{2}$
- C. sqrt2$\sqrt{2}$
- D. 2
Solution
### Related Formula
Leibniz Integral Rule for differentiation under integral sign:
fracmathrmdmathrmdxleft(int_phi(x)^psi(x) f(t)mathrmdtright) = f(psi(x))psi'(x) - f(phi(x))phi'(x)$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{\phi(x)}^{\psi(x)} f(t)\mathrm{d}t\right) = f(\psi(x))\psi'(x) - f(\phi(x))\phi'(x)$$
### Core Logic
Rewrite equation to isolate the integral kernel:
y = 1 - 2x + e^x int_0^x e^-t f(t)mathrmdt$$y = 1 - 2x + e^x \int_0^x e^{-t} f(t)\mathrm{d}t$$
Differentiating with respect to x$x$ using product rule and Leibniz rule:
fracmathrmdymathrmdx = -2 + e^x int_0^x e^-t f(t)mathrmdt + e^x cdot left(e^-x f(x)right)$$\frac{\mathrm{d}y}{\mathrm{d}x} = -2 + e^x \int_0^x e^{-t} f(t)\mathrm{d}t + e^x \cdot \left(e^{-x} f(x)\right)$$
Notice that e^x int_0^x e^-t f(t)mathrmdt = y - (1 - 2x)$e^x \int_0^x e^{-t} f(t)\mathrm{d}t = y - (1 - 2x)$. Substitute this back:
fracmathrmdymathrmdx = -2 + [y - 1 + 2x] + y implies fracmathrmdymathrmdx - 2y = 2x - 3$$\frac{\mathrm{d}y}{\mathrm{d}x} = -2 + [y - 1 + 2x] + y \implies \frac{\mathrm{d}y}{\mathrm{d}x} - 2y = 2x - 3$$
### Step 1: Solve Differential Equation
Integrating factor textI.F. = e^int -2 mathrmdx = e^-2x$\text{I.F.} = e^{\int -2 \mathrm{d}x} = e^{-2x}$.
y e^-2x = int (2x - 3)e^-2xmathrmdx = frac-(2x - 3)2e^-2x - frac12e^-2x + c$$y e^{-2x} = \int (2x - 3)e^{-2x}\mathrm{d}x = \frac{-(2x - 3)}{2}e^{-2x} - \frac{1}{2}e^{-2x} + c$$
From original equation, at x=0$x=0$, f(0) = 1$f(0) = 1$. Evaluating c$c$:
1 = frac32 - frac12 + c implies c = 0$$1 = \frac{3}{2} - \frac{1}{2} + c \implies c = 0$$
Thus, y = -x + 1 implies x + y = 1$y = -x + 1 \implies x + y = 1$.
### Step 2: Calculate Area
The boundary line is x + y = 1$x + y = 1$. The area bounded by this line and coordinate axes is a right triangle with intercepts (1,0)$(1,0)$ and (0,1)$(0,1)$:
textArea = frac12 times 1 times 1 = frac12$$\text{Area} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$$
### Pattern Recognition
Integral equations of convolution type (e^x-t$e^{x-t}$) always simplify directly into standard linear ordinary differential equations of first or second order when tracking Leibniz rules properly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Differential Equations
Class 12 Mathematics: Area Under Curves
Q54
jee_main_2025_24_jan_evening
Linear Differential Equations
Let f:(0,infty)rightarrow R$f:(0,\infty)\rightarrow R$ be a function which is differentiable at all points of its domain and satisfies the condition x^2f^prime(x)=2xf(x)+3,$x^{2}f^{\prime}(x)=2xf(x)+3,$ with f(1)=4$f(1)=4$ Then 2f(2)$2f(2)$ is equal to: [cite: 3263, 3264]
- A. 29$29$
- B. 19$19$
- C. 39$39$
- D. 23$23$
Solution
### Related Formula
The quotient rule derivative identity is given by:
fracddxleft(fracf(x)x^2right) = fracx^2 f'(x) - 2x f(x)x^4$$\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = \frac{x^2 f'(x) - 2x f(x)}{x^4}$$
### Core Logic
Rearrange the given differential condition:
x^2 f'(x) - 2x f(x) = 3$$x^2 f'(x) - 2x f(x) = 3$$
### Step 1: Divide by x^4$x^4$
To convert the left-hand side into an exact derivative form, divide the full relation by x^4$x^4$:
fracx^2 f'(x) - 2x f(x)x^4 = frac3x^4$$\frac{x^2 f'(x) - 2x f(x)}{x^4} = \frac{3}{x^4}$$
fracddxleft(fracf(x)x^2right) = 3x^-4$$\frac{d}{dx}\left(\frac{f(x)}{x^2}\right) = 3x^{-4}$$
### Step 2: Integration and Evaluating Constant
Integrating both sides with respect to x$x$ [cite: 3893, 3894]:
fracf(x)x^2 = int 3x^-4 dx = -x^-3 + C = -frac1x^3 + C$$\frac{f(x)}{x^2} = \int 3x^{-4} dx = -x^{-3} + C = -\frac{1}{x^3} + C$$
f(x) = -frac1x + Cx^2$$f(x) = -\frac{1}{x} + Cx^2$$
Using the given value f(1) = 4$f(1) = 4$ [cite: 3264, 3896]:
4 = -frac11 + C(1)^2 Rightarrow 4 = -1 + C Rightarrow C = 5$$4 = -\frac{1}{1} + C(1)^2 \Rightarrow 4 = -1 + C \Rightarrow C = 5$$
Thus, the function is f(x) = -frac1x + 5x^2$f(x) = -\frac{1}{x} + 5x^2$.
### Step 3: Calculating 2f(2)
Substitute $2f(2)
Substitute $x = 2 to compute $ to compute $2f(2) :
$ :
$2 times f(2) = 2 times left[ -frac12 + 5(2)^2 right]$2 \times f(2) = 2 \times \left[ -\frac{1}{2} + 5(2)^2 \right]$
$
$2 times f(2) = 2 times left[ -frac12 + 20 right] = -1 + 40 = 39$2 \times f(2) = 2 \times \left[ -\frac{1}{2} + 20 \right] = -1 + 40 = 39$ [cite: 3900, 3901]
### Pattern Recognition
Recognizing the structure $ [cite: 3900, 3901]
### Pattern Recognition
Recognizing the structure $x^2 f'(x) - 2x f(x) as a partial quotient rule is faster than formatting it into standard linear order $ as a \partial quotient rule is faster than formatting it into standard linear order $\frac{dy}{dx} + P(x)y = Q(x)$ format, though both methods lead to the identical integration parameters safely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Differential Equations
