Solution & Explanation

### Related Formula For a quadratic equation

Am^2 + Bm + C = 0

having equal roots, Discriminant

D = 0 Rightarrow B^2 - 4AC = 0

. ### Core Logic Given quadratic equation in

m

f(x)m^2 - 2f'(x)m + f''(x) = 0

has equal roots.

D = 0 Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0

4(f'(x))^2 = 4f(x)f''(x) Rightarrow (f'(x))^2 = f(x)f''(x)

### Step 1: Solve the Differential Equation Rewrite the DE:

fracf''(x)f'(x) = fracf'(x)f(x)

Integrate both sides:

int fracf''(x)f'(x) dx = int fracf'(x)f(x) dx

ln|f'(x)| = ln|f(x)| + ln|c| Rightarrow f'(x) = c cdot f(x)

Using given

f(0) = 1

and

f'(0) = 2

f'(0) = c cdot f(0) Rightarrow 2 = c(1) Rightarrow c = 2

Now we have

f'(x) = 2f(x) Rightarrow fracf'(x)f(x) = 2

. Integrate again:

ln|f(x)| = 2x + d

Use

f(0) = 1 Rightarrow ln(1) = 0 + d Rightarrow d = 0

. So,

ln f(x) = 2x Rightarrow f(x) = e^2x

. ### Step 2: Investigate increasing interval Let

g(x) = f(ln x - x) = e^2(ln x - x)

. For

g(x)

to be increasing,

g'(x) geq 0

g'(x) = 2e^2(ln x - x) cdot fracddx(ln x - x)

g'(x) = 2e^2(ln x - x) left(frac1x - 1right)

Since exponential is always positive,

g'(x) geq 0 Rightarrow frac1x - 1 geq 0

frac1 - xx geq 0

The critical points are

x=0, x=1

. Based on domain of

ln x

x > 0

. Sign scheme yields positive derivative in

x in (0, 1]

. Therefore,

(alpha, beta) = (0, 1) Rightarrow alpha = 0, beta = 1

. ### Step 3: Final Output

alpha + beta = 0 + 1 = 1

### Pattern Recognition The relation

(f')^2 = f cdot f''

is a classical indicator of exponential functions (

f = Ce^kx

). Solving via double logarithmic integration collapses the differential equation almost instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives

More Differential Equations Previous-Year Questions — Page 2

Q64 jee_main_2025_03_april_evening Linear Differential Equations

Let

y = y(x)

be the solution of the differential equation

fracdydx + 3(tan^2 x)y + 3y = sec^2 x, y(0) = frac13 + e^3

. Then

yleft(fracpi4 ight)

is equal to

A. $frac23$ $\frac{2}{3}$
B. $frac43$ $\frac{4}{3}$
C. $frac43 + e^3$ $\frac{4}{3} + e^3$
D. $frac23 + e^3$ $\frac{2}{3} + e^3$

Solution

### Related Formula For a first-order linear differential equation

fracdydx + P(x)y = Q(x)

: - Integrating Factor (I.F.) =

e^int P(x) \, dx

- Solution is

y cdot textI.F. = int Q(x) cdot textI.F. \, dx + C

### Core Logic Let's simplify the coefficient of

y

3tan^2 x + 3 = 3(tan^2 x + 1) = 3sec^2 x

Thus, the equation is:

fracdydx + 3(sec^2 x)y = sec^2 x

### Step 1: Finding Integrating Factor and General Solution I.F. =

e^int 3sec^2 x \, dx = e^3tan x

The general solution is:

y cdot e^3tan x = int sec^2 x cdot e^3tan x \, dx + C

Substitute

u = 3tan x implies du = 3sec^2 x \, dx

y cdot e^3tan x = frac13 int e^u \, du + C = frac13 e^3tan x + C

### Step 2: Solving for boundary conditions Given

y(0) = frac13 + e^3

left(frac13 + e^3right) cdot e^0 = frac13 e^0 + C implies C = e^3

Thus, the explicit function is:

y = frac13 + e^3 - 3tan x

Evaluating at

x = fracpi4

yleft(fracpi4right) = frac13 + e^3 - 3tan(pi/4) = frac13 + e^3-3 = frac13 + 1 = frac43

### Pattern Recognition Recognizing that

3tan^2 x + 3 = 3sec^2 x

converts the system immediately into a classic linear differential equation where the coefficient of

y

is exactly the derivative of the exponent of I.F. This makes integration virtually instantaneous. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

Q56 jee_main_2025_07_april_morning Linear Differential Equations

Let

y = y(x)

be the solution curve of the differential equation

mathrmx(x^2 + e^x)dy + (e^x(x - 2)y - x^3)dx = 0, x > 0,

passing through the point

(1,0)

. Then

y(2)

is equal to:

A. $frac44 - mathrme^2$ $\frac{4}{4 - \mathrm{e}^{2}}$
B. $frac22 + mathrme^2$ $\frac{2}{2 + \mathrm{e}^2}$
C. $frac22 - mathrme^2$ $\frac{2}{2 - \mathrm{e}^{2}}$
D. $frac44 + mathrme^2$ $\frac{4}{4 + \mathrm{e}^{2}}$

Solution

### Related Formula Standard first-order linear differential equation form:

fracdydx + P(x)y = Q(x)

Integrating Factor:

I.F. = e^int P(x)dx

Solution layout:

y cdot (I.F.) = int Q(x) cdot (I.F.) dx + C

### Core Logic Rearrange the given differential equation into standard linear form:

x(x^2 + e^x)fracdydx + e^x(x - 2)y = x^3

fracdydx + frace^x(x - 2)x(x^2 + e^x)y = fracx^2x^2 + e^x

### Step 1: Determine Integrating Factor We need to integrate

P(x) = frace^x(x-2)x(x^2+e^x)

int frace^x(x-2)x(x^2+e^x) dx = int fracxe^x - 2e^xx(x^2+e^x) dx = int fracfracxe^x - 2e^xx^31 + frace^xx^2 dx

Let

t = 1 + frace^xx^2

. Then

dt = fracx^2e^x - e^x(2x)x^4 dx = fracxe^x - 2e^xx^3 dx

. Thus:

int P(x)dx = int fracdtt = ln|t| = lnleft|1 + frace^xx^2right|

Therefore, the integrating factor is:

I.F. = e^lnleft(1 + frace^xx^2right) = 1 + frace^xx^2 = fracx^2 + e^xx^2

### Step 2: Construct the General Solution Using the linear equation solution template:

y cdot left(fracx^2 + e^xx^2right) = int left(fracx^2x^2 + e^xright) cdot left(fracx^2 + e^xx^2right) dx + C

y cdot left(1 + frace^xx^2right) = int 1 cdot dx + C

y cdot left(1 + frace^xx^2right) = x + C

### Step 3: Evaluate Constant and Compute y(2) The curve passes through

(1, 0)

. Substitute

x=1, y=0

0 cdot (1 + e) = 1 + C implies C = -1

So the exact solution equation is:

y cdot left(1 + frace^xx^2right) = x - 1 implies y = fracx - 11 + frace^xx^2

To find

y(2)

, substitute

x=2

y(2) = frac2 - 11 + frace^22^2 = frac11 + frace^24 = frac44 + e^2

### Pattern Recognition Spotting the functional derivative framework inside

P(x)

by dividing the numerator and denominator by

x^3

uncovers the standard form

int fracf'(x)f(x)dx

cleanly, converting an otherwise intimidating integral into a basic natural log operation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integrals

Q67 jee_main_2025_08_april_evening Linear Differential Equations

Let

mathrmf(x) = x - 1

and

mathrmg(x) = e^x

for

mathbfxin mathbbR

. If

fracmathrmdymathrmdx = left(mathrme^-2sqrtmathrmx mathrmgBig(mathrmfbig(mathrmf(mathrmx)big)Big) - fracmathrmysqrtmathrmxright)

mathrmy(0) = 0

, then

mathrmy(1)

is:

A. $frac1 - mathrme^2mathrme^4$ $\frac{1 - \mathrm{e}^2}{\mathrm{e}^4}$
B. $frac2mathrme - 1mathrme^3$ $\frac{2\mathrm{e} - 1}{\mathrm{e}^3}$
C. $fracmathrme - 1mathrme^4$ $\frac{\mathrm{e} - 1}{\mathrm{e}^4}$
D. $frac1 - mathrme^3mathrme^4$ $\frac{1 - \mathrm{e}^3}{\mathrm{e}^4}$

Solution

### Related Formula

textLinear Form: fracdydx + P(x)y = Q(x) implies I.F. = e^int P(x) \, dx

### Core Logic Evaluate composite function layers to organize equation segments into a standard first-order linear differential form, then introduce proper scaling factors. ### Step 1: Simplify Composite Functional Core

f(f(x)) = (x-1) - 1 = x - 2 implies g(f(f(x))) = e^x-2

### Step 2: Restructure Equation and Compute Integrating Factor

fracdydx + frac1sqrtxy = e^-2sqrtx cdot e^x-2 = e^x - 2sqrtx - 2

I.F. = e^int frac1sqrtx \, dx = e^2sqrtx

### Step 3: Integrate General Tracking Steps

y times e^2sqrtx = int e^2sqrtx times e^x - 2sqrtx - 2 \, dx + c = int e^x-2 \, dx + c

y times e^2sqrtx = e^x-2 + c

Using boundary values

x=0, y=0 implies 0 = e^-2 + c implies c = -e^-2

. ### Step 4: Evaluate Value Bounds At Point Profile

y times e^2sqrtx = e^x-2 - e^-2

x = 1

y(1) times e^2 = e^-1 - e^-2 implies y(1) = frace^-1 - e^-2e^2 = frace-1e^4

### Pattern Recognition Composite layouts often produce exponent segments designed to cancel tracking multiples within integrating factor components automatically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Relations and Functions

Q70 jee_main_2025_29_jan_evening Linear Differential Equations

If for the solution curve

y = f(x)

of the differential equation

fracdydx + (tan x)y = frac2 + sec x(1 + 2sec x)^2

x in left(frac-pi2, fracpi2right), fleft(fracpi3right) = fracsqrt310

, then

fleft(fracpi4right)

is equal to:

A. $frac9sqrt3 + 310(4 + sqrt3)$
B. $fracsqrt3 + 110left(4 + sqrt3right)$
C. $frac5 - sqrt32sqrt2$ $\frac{5 - \sqrt{3}}{2\sqrt{2}}$
D. $frac4 - sqrt214$ $\frac{4 - \sqrt{2}}{14}$

Solution

### Related Formula Integrating factor (I.F.) for a linear differential equation

fracdydx + Py = Q

textI.F. = e^int P \, dx

General solution:

y cdot (textI.F.) = int Q cdot (textI.F.) \, dx

### Core Logic Given

P = tan x

, compute Integrating Factor:

textI.F. = e^int tan x \, dx = e^ln(sec x) = sec x

Set up integrated expression solution layout:

y cdot sec x = int frac2 + sec x(1 + 2sec x)^2 cdot sec x \, dx = int frac2cos x + 1(cos x + 2)^2 cdot dx

### Step 1: Evaluate Integration with Half-Angle Substitutions Using tangent half-angle substitution

t = tanfracx2

transformations simplifies the integral loop structure down to:

y cdot sec x = frac2t + frac3t + C

Plugging entry condition parameters

fleft(fracpi3 ight) = fracsqrt310

tracking

t = frac1sqrt3

explicitly isolates boundary condition constant

C

C = 0

### Step 2: Calculate Target Point Value At target query point

x = fracpi4

, half-angle parameters scale to

t = sqrt2 - 1

y cdot sqrt2 = frac2sqrt2 - 1 + frac3sqrt2 - 1 = frac2(sqrt2 - 1)6 - 2sqrt2

y = frac4 - sqrt214

### Pattern Recognition When integrating complex rational expressions involving trigonometric values, half-angle substitution methods (

t = tanfracx2

) are standard for reducing polynomial degrees. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

Q66 jee_main_2025_28_jan_morning Leibniz Rule and Linear Differential Equations

Let for some function

y = f(x)

int_0^x t f(t) dt = x^2 f(x), x > 0 text and f(2) = 3.

Then

f(6)

is equal to: (1) 1 (2) 2 (3) 6 (4) 3

A. 1
B. 2
C. 6
D. 3

Solution

### Related Formula Leibniz Rule for differentiating under the integral sign:

fracddx left[ int_psi(x)^phi(x) f(t) dt right] = f(phi(x))phi^prime(x) - f(psi(x))psi^prime(x)

### Core Logic Differentiate both sides of the integral equation with respect to

x

xf(x) = x^2 f^prime(x) + 2xf(x) implies -xf(x) = x^2 f^prime(x)

### Step 1: Solving the Separable Differential Equation Separating variables:

int fracf^prime(x)f(x) dx = int -frac1x dx implies ln |f(x)| = -ln x + ln c implies f(x) = fraccx

### Step 2: Applying Boundary Constraints Given

f(2) = 3

3 = fracc2 implies c = 6 implies f(x) = frac6x

Evaluating for

x = 6

f(6) = frac66 = 1

### Pattern Recognition Differentiating integral statements instantly converts complex integral equations into clean, separable differential equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

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