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 7
Q27
jee_main_2024_30_jan_morning
Linear Differential Equations
Let
y = y(x)$y = y(x)$ be the
solution of the differential equation (1 - x^2) dy = left[ xy + (x^3 + 2)sqrt3(1 - x^2) right] dx, quad -1 < x < 1, y(0) = 0$(1 - x^2) dy = \left[ xy + (x^3 + 2)\sqrt{3(1 - x^2)} \right] dx, \quad -1 < x < 1, y(0) = 0$. If
yleft(frac12right) = fracmn$y\left(\frac{1}{2}\right) = \frac{m}{n}$,
m$m$ and
n$n$ are co-prime numbers, then
m + n$m + n$ is equal to
Numerical Answer. Answer: 97 to 97
Solution
### Related Formula
For a linear differential equation fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$:
IF = e^int P(x) dx$$IF = e^{\int P(x) dx}$$
Solution: y cdot IF = int Q(x) cdot IF \, dx + C$y \cdot IF = \int Q(x) \cdot IF \, dx + C$
### Core Logic
Rearrange the given differential equation into standard linear form:
(1 - x^2) fracdydx = xy + (x^3 + 2)sqrt3(1 - x^2)$$(1 - x^2) \frac{dy}{dx} = xy + (x^3 + 2)\sqrt{3(1 - x^2)}$$
fracdydx - fracx1 - x^2 y = frac(x^3 + 2)sqrt3(1 - x^2)1 - x^2$$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{(x^3 + 2)\sqrt{3(1 - x^2)}}{1 - x^2}$$
fracdydx - fracx1 - x^2 y = fracsqrt3(x^3 + 2)sqrt1 - x^2$$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}}$$
Identify P(x) = -fracx1 - x^2$P(x) = -\frac{x}{1 - x^2}$.
### Step 1: Finding Integrating Factor (IF)
IF = e^int -fracx1 - x^2 dx$$IF = e^{\int -\frac{x}{1 - x^2} dx}$$
Let 1 - x^2 = t Rightarrow -2x dx = dt Rightarrow -x dx = fracdt2$1 - x^2 = t \Rightarrow -2x dx = dt \Rightarrow -x dx = \frac{dt}{2}$.
IF = e^frac12 int frac1t dt = e^frac12 ln t = e^ln sqrtt = sqrt1 - x^2$$IF = e^{\frac{1}{2} \int \frac{1}{t} dt} = e^{\frac{1}{2} \ln t} = e^{\ln \sqrt{t}} = \sqrt{1 - x^2}$$
### Step 2: General Solution
The solution is given by:
y sqrt1 - x^2 = int left( fracsqrt3(x^3 + 2)sqrt1 - x^2 right) sqrt1 - x^2 \, dx + C$$y \sqrt{1 - x^2} = \int \left( \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}} \right) \sqrt{1 - x^2} \, dx + C$$
y sqrt1 - x^2 = sqrt3 int (x^3 + 2) \, dx + C$$y \sqrt{1 - x^2} = \sqrt{3} \int (x^3 + 2) \, dx + C$$
y sqrt1 - x^2 = sqrt3 left( fracx^44 + 2x right) + C$$y \sqrt{1 - x^2} = \sqrt{3} \left( \frac{x^4}{4} + 2x \right) + C$$
### Step 3: Finding Constant C
Using y(0) = 0$y(0) = 0$:
0 cdot 1 = sqrt3(0 + 0) + C Rightarrow C = 0$$0 \cdot 1 = \sqrt{3}(0 + 0) + C \Rightarrow C = 0$$
So, y(x) = fracsqrt3sqrt1 - x^2 left( fracx^44 + 2x right)$y(x) = \frac{\sqrt{3}}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x \right)$.
### Step 4: Evaluating required point
Substitute x = 1/2$x = 1/2$:
yleft(frac12right) = fracsqrt3sqrt1 - 1/4 left( frac1/164 + 2left(frac12right) right)$$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{1 - 1/4}} \left( \frac{1/16}{4} + 2\left(\frac{1}{2}\right) \right)$$
Wait, frac(1/2)^44 = frac1/164 = frac164$\frac{(1/2)^4}{4} = \frac{1/16}{4} = \frac{1}{64}$. Let's recheck the expression:
yleft(frac12right) = fracsqrt3sqrt3/4 left( frac164 + 1 right)$$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{3/4}} \left( \frac{1}{64} + 1 \right)$$
= fracsqrt3sqrt3/2 left( frac6564 right) = 2 times frac6564 = frac6532$$= \frac{\sqrt{3}}{\sqrt{3}/2} \left( \frac{65}{64} \right) = 2 \times \frac{65}{64} = \frac{65}{32}$$
Here, m = 65$m = 65$ and n = 32$n = 32$. They are co-prime.
Thus, m + n = 65 + 32 = 97$m + n = 65 + 32 = 97$.
### Pattern Recognition
Whenever roots matching the integration denominator appear on the RHS of a linear differential setup, it is a high-confidence signal that the integrating factor cleanly annihilates the fractional root component during the solution stage.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Differential Equations
Q8
jee_main_2024_31_jan_evening
Newton's Law of Cooling
The
temperature T(t)$T(t)$ of a body at time
t = 0$t = 0$ is
160^circ mathrmF$160^{\circ} \mathrm{F}$ and it decreases continuously as per the differential equation
fracmathrmdTmathrmdt = -mathrmK(T - 80)$\frac{\mathrm{dT}}{\mathrm{dt}} = -\mathrm{K(T - 80)}$, where
K$K$ is positive constant. If
T(15) = 120^circmathrmF$T(15) = 120^{\circ}\mathrm{F}$, then
T(45)$T(45)$ is equal to
- A. 85^circmathrmF$85^{\circ}\mathrm{F}$
- B. 95^circmathrmF$95^{\circ}\mathrm{F}$
- C. 90^mathrmomathrmF$90^{\mathrm{o}}\mathrm{F}$
- D. 80^circmathrmF$80^{\circ}\mathrm{F}$
Solution
### Related Formula
int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C$$\int \frac{dT}{T-T_s} = -K \int dt \implies \ln|T-T_s| = -Kt + C$$
### Core Logic
Given fracdTdt = -K(T-80)$\frac{dT}{dt} = -K(T-80)$.
Integrating from t=0$t=0$ to t$t$:
int_160^T fracdTT-80 = -K int_0^t dt$$\int_{160}^{T} \frac{dT}{T-80} = -K \int_0^t dt$$
[ln|T-80|]_160^T = -Kt$$[\ln|T-80|]_{160}^T = -Kt$$
lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt$$\ln\left(\frac{T-80}{80}\right) = -Kt \implies T(t) = 80 + 80e^{-Kt}$$
Given T(15) = 120$T(15) = 120$:
120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12$$120 = 80 + 80e^{-15K} \implies 40 = 80e^{-15K} \implies e^{-15K} = \frac{1}{2}$$
To find T(45)$T(45)$:
T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3$$T(45) = 80 + 80e^{-45K} = 80 + 80(e^{-15K})^3$$
= 80 + 80left(frac12right)^3$$= 80 + 80\left(\frac{1}{2}\right)^3$$
= 80 + 80left(frac18right) = 90$$= 80 + 80\left(\frac{1}{8}\right) = 90$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Differential Equations
Q29
jee_main_2024_31_jan_evening
Linear Differential Equations
Let
y = y(x)$y = y(x)$ be the solution of the
differential equation sec^2 x \, dx + left( e^2y tan^2 x + tan x right) dy = 0$\sec^2 x \, dx + \left( e^{2y} \tan^2 x + \tan x \right) dy = 0$,
0 < x < fracpi2$0 < x < \frac{\pi}{2}$,
yleft(fracpi4right) = 0$y\left(\frac{\pi}{4}\right) = 0$. If
yleft(fracpi6right) = alpha$y\left(\frac{\pi}{6}\right) = \alpha$. Then
e^8alpha$e^{8\alpha}$ is equal to
Numerical Answer. Answer: 9 to 9
Solution
### Related Formula
textIntegrating factor for fracdudy + P(y)u = Q(y) text is I.F. = e^int P(y)dy$$\text{Integrating factor for } \frac{du}{dy} + P(y)u = Q(y) \text{ is } I.F. = e^{\int P(y)dy}$$
### Core Logic
Given DE: sec^2 x fracdxdy + e^2y tan^2 x + tan x = 0$\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$
Substitute t = tan x implies fracdtdy = sec^2 x fracdxdy$t = \tan x \implies \frac{dt}{dy} = \sec^2 x \frac{dx}{dy}$.
The DE becomes:
fracdtdy + t = -t^2 e^2y$$\frac{dt}{dy} + t = -t^2 e^{2y}$$
This is a Bernoulli equation in t$t$. Divide by t^2$t^2$:
frac1t^2fracdtdy + frac1t = -e^2y$$\frac{1}{t^2}\frac{dt}{dy} + \frac{1}{t} = -e^{2y}$$
Substitute u = frac1t implies fracdudy = -frac1t^2fracdtdy$u = \frac{1}{t} \implies \frac{du}{dy} = -\frac{1}{t^2}\frac{dt}{dy}$.
-fracdudy + u = -e^2y implies fracdudy - u = e^2y$$-\frac{du}{dy} + u = -e^{2y} \implies \frac{du}{dy} - u = e^{2y}$$
This is a linear DE in u$u$ with respect to y$y$. P(y) = -1, Q(y) = e^2y$P(y) = -1, Q(y) = e^{2y}$.
Integrating Factor: I.F. = e^int -1 dy = e^-y$I.F. = e^{\int -1 dy} = e^{-y}$.
Solution:
u e^-y = int e^2y e^-y dy = int e^y dy = e^y + C$$u e^{-y} = \int e^{2y} e^{-y} dy = \int e^y dy = e^y + C$$
Substitute u = frac1tan x$u = \frac{1}{\tan x}$:
frace^-ytan x = e^y + C$$\frac{e^{-y}}{\tan x} = e^y + C$$
Use given condition y(pi/4) = 0$y(\pi/4) = 0$:
frace^0tan(pi/4) = e^0 + C implies 1 = 1 + C implies C = 0$$\frac{e^0}{\tan(\pi/4)} = e^0 + C \implies 1 = 1 + C \implies C = 0$$
Therefore, frace^-ytan x = e^y implies tan x = e^-2y$\frac{e^{-y}}{\tan x} = e^y \implies \tan x = e^{-2y}$.
Evaluate at x = pi/6$x = \pi/6$, y = alpha$y = \alpha$:
tan(pi/6) = e^-2alpha implies frac1sqrt3 = e^-2alpha$$\tan(\pi/6) = e^{-2\alpha} \implies \frac{1}{\sqrt{3}} = e^{-2\alpha}$$
e^2alpha = sqrt3 implies (e^2alpha)^4 = (sqrt3)^4 = 9$$e^{2\alpha} = \sqrt{3} \implies (e^{2\alpha})^4 = (\sqrt{3})^4 = 9$$
Thus, e^8alpha = 9$e^{8\alpha} = 9$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Differential Equations
Q9
jee_main_2024_31_jan_morning
Homogeneous Differential Equations
The solution curve of the differential equation yfracdxdy = x(log_e x - log_e y + 1), x > 0, y > 0$y\frac{dx}{dy} = x(\log_e x - \log_e y + 1), x > 0, y > 0$ passing through the point (e, 1)$(e, 1)$ is
- A. |log_e fracyx| = x$$|\log_e \frac{y}{x}| = x$$
- B. |log_e fracyx| = y^2$$|\log_e \frac{y}{x}| = y^2$$
- C. |log_e fracxy| = y$$|\log_e \frac{x}{y}| = y$$
- D. 2|log_e fracxy| = y + 1$$2|\log_e \frac{x}{y}| = y + 1$$
Solution
### Core Logic
Given DE: fracdxdy = fracxy left(lnleft(fracxyright) + 1right)$\frac{dx}{dy} = \frac{x}{y} \left(\ln\left(\frac{x}{y}\right) + 1\right)$
Let fracxy = t implies x = ty$\frac{x}{y} = t \implies x = ty$.
Differentiating w.r.t y$y$:
fracdxdy = t + yfracdtdy$$\frac{dx}{dy} = t + y\frac{dt}{dy}$$
### Step 1: Substitution and Integration
t + yfracdtdy = t(ln(t) + 1) = tln t + t$$t + y\frac{dt}{dy} = t(\ln(t) + 1) = t\ln t + t$$
yfracdtdy = tln t implies fracdttln t = fracdyy$$y\frac{dt}{dy} = t\ln t \implies \frac{dt}{t\ln t} = \frac{dy}{y}$$
Integrate both sides. Let ln t = p implies frac1t dt = dp$\ln t = p \implies \frac{1}{t} dt = dp$.
int fracdpp = int fracdyy$$\int \frac{dp}{p} = \int \frac{dy}{y}$$
ln|p| = ln y + C implies ln|ln t| = ln y + C$$\ln|p| = \ln y + C \implies \ln|\ln t| = \ln y + C$$
lnleft|lnleft(fracxyright)right| = ln y + C$$\ln\left|\ln\left(\frac{x}{y}\right)\right| = \ln y + C$$
### Step 2: Applying Boundary Conditions
Given curve passes through (e, 1)$(e, 1)$:
lnleft|lnleft(frace1right)right| = ln(1) + C implies C = 0$$\ln\left|\ln\left(\frac{e}{1}\right)\right| = \ln(1) + C \implies C = 0$$
lnleft|lnleft(fracxyright)right| = ln y$$\ln\left|\ln\left(\frac{x}{y}\right)\right| = \ln y$$
left|lnleft(fracxyright)right| = e^ln y = y$$\left|\ln\left(\frac{x}{y}\right)\right| = e^{\ln y} = y$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Differential Equations
Q11
jee_main_2024_31_jan_morning
Linear Differential Equations
Let y = y(x)$y = y(x)$ be the solution of the differential equation fracdydx = frac(tan x) + ysin x(sec x - sin x tan x), x in left(0, fracpi2right)$\frac{dy}{dx} = \frac{(\tan x) + y}{\sin x(\sec x - \sin x \tan x)}, x \in \left(0, \frac{\pi}{2}\right)$ satisfying the condition yleft(fracpi4right) = 2$y\left(\frac{\pi}{4}\right) = 2$. Then, yleft(fracpi3right)$y\left(\frac{\pi}{3}\right)$ is
- A. sqrt3(2 + log_esqrt3)$$\sqrt{3}(2 + \log_e\sqrt{3})$$
- B. fracsqrt32(2 + log_e 3)$$\frac{\sqrt{3}}{2}(2 + \log_e 3)$$
- C. sqrt3(1 + 2log_e 3)$$\sqrt{3}(1 + 2\log_e 3)$$
- D. sqrt3(2 + log_e 3)$$\sqrt{3}(2 + \log_e 3)$$
Solution
### Core Logic
fracdydx = fracfracsin xcos x + ysin x left(frac1cos x - fracsin^2 xcos xright) = fracsin x + ycos xsin x (1 - sin^2 x)$$\frac{dy}{dx} = \frac{\frac{\sin x}{\cos x} + y}{\sin x \left(\frac{1}{\cos x} - \frac{\sin^2 x}{\cos x}\right)} = \frac{\sin x + y\cos x}{\sin x (1 - \sin^2 x)}$$
fracdydx = fracsin x + ycos xsin x cos^2 x = sec^2 x + frac2ysin 2x$$\frac{dy}{dx} = \frac{\sin x + y\cos x}{\sin x \cos^2 x} = \sec^2 x + \frac{2y}{\sin 2x}$$
fracdydx - 2csc(2x)y = sec^2 x$$\frac{dy}{dx} - 2\csc(2x)y = \sec^2 x$$
### Step 1:
Integrating Factor
This is an LDE of form
fracdydx + Py = Q$\frac{dy}{dx} + Py = Q$.
I.F. = e^int -2csc(2x) dx$$I.F. = e^{\int -2\csc(2x) dx}$$
Let
2x = t implies 2dx = dt$2x = t \implies 2dx = dt$.
I.F. = e^-int csc t dt = e^-ln|tan(t/2)| = e^-ln|tan x| = frac1|tan x|$$I.F. = e^{-\int \csc t dt} = e^{-\ln|\tan(t/2)|} = e^{-\ln|\tan x|} = \frac{1}{|\tan x|}$$
### Step 2: Solution of LDE
y(I.F.) = int Q(I.F.) dx + C$$y(I.F.) = \int Q(I.F.) dx + C$$
yfrac1tan x = int sec^2 x frac1tan x dx + C$$y\frac{1}{\tan x} = \int \sec^2 x \frac{1}{\tan x} dx + C$$
Let
tan x = t implies sec^2 x dx = dt$\tan x = t \implies \sec^2 x dx = dt$.
yfrac1tan x = int fracdtt + C = ln|tan x| + C$$y\frac{1}{\tan x} = \int \frac{dt}{t} + C = \ln|\tan x| + C$$
y = tan x(ln|tan x| + C)$$y = \tan x(\ln|\tan x| + C)$$
### Step 3: Boundary Value
Given
y(pi/4) = 2$y(\pi/4) = 2$:
2 = 1(ln 1 + C) implies C = 2$$2 = 1(\ln 1 + C) \implies C = 2$$
Thus,
y = tan x (ln|tan x| + 2)$y = \tan x (\ln|\tan x| + 2)$.
At
x = pi/3$x = \pi/3$:
y(pi/3) = sqrt3(lnsqrt3 + 2)$$y(\pi/3) = \sqrt{3}(\ln\sqrt{3} + 2)$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Differential Equations
