Find a particular solution of the differential equation (x + 1) dy/dx = 2 e⁻ʸ − 1, given that y = 0 when x = 0

 Find a particular solution of the differential equation

(x + 1) dy/dx = 2 e⁻ʸ − 1,

given that y = 0 when x = 0

variable separable differential equation with initial condition

given

(x + 1) dy/dx = 2 e⁻ʸ − 1,

separating

 dy / (2 e⁻ʸ − 1) = dx / (x + 1)

multiply numerator and denominator of LHS with  eʸ 

eʸ dy / (2 − eʸ) = dx / (x + 1)

integrating

∫ eʸ / (2 − eʸ) dy = ∫ 1 / (x + 1) dx

for the LHS use the substitution

u = 2 − eʸ 

⇒ du = −eʸ dy

eʸ dy =-du

we get

 ∫ −du / u = ∫ 1 / (x + 1)  dx

 −ln|u| = ln|x + 1| + C

re substitute the value of u = 2 − eʸ 

 −ln|2 − eʸ| = ln|x + 1| + C

 ln|2 − eʸ| +ln|x + 1| = -C

using properties of logarithms,

ln|(2 − eʸ)(x + 1)| = -C

using definition of logarithm

 (x + 1)(2 − eʸ) = A-----(1)

, where A is another arbitrary constant

use the initial condition   y = 0 when x = 0 to solve for A

we get A =1

(1)⇒ (x + 1)(2 − eʸ) =1

or

2 − eʸ = 1 / (x + 1)

eʸ = 2 − [ 1 / (x + 1)]

eʸ =  (2x + 1) / (x + 1)

using definition of log

y = ln[(2x + 1) / (x + 1)]

for more details watch this video 

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variable separable differential equation dy/dx = (2 − y) / (x + 1)

 Solve the following variable separable differential equation

dy/dx = (2 − y) / (x + 1)

dy/dx = (2 − y) / (x + 1)

clearly it is variable separable type

dy / (2 − y) = dx / (x + 1)

integrating both sides

∫ dy / (2 − y) = ∫ dx / (x + 1)

− ln|2 − y| = ln|x + 1| + lnC [ note the ( -1), coefficient of y when integrating]

0 = ln|2 − y| + ln|x + 1| + lnC

use the property of logarithms

0=ln [ |2 − y| |x + 1| C ]

or

[ |2 − y| |x + 1| C ] = 1

for more details refer to this video

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variable sepaprable differential equation

Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

 Three consecutive positive integers are such that sum of square of the first and the product of the other two is 67, find the integers.

Let the three consecutive positive integers be n, n+1, n+2

Given that

that sum of square of the first and the product of the other two is 67

n² + (n+1)(n+2) = 67

n² + (n² + 2n + n + 2) = 67  

n² + n² + 3n + 2 = 67  

2n² + 3n + 2 = 67  

2n² + 3n + 2 - 67  = 0

2n² + 3n − 65 = 0

Solve by quadratic formula or factorisation

Factorisation method

Find two numbers that gives a product of 2 × (−65) = −130 and add to 3.

by trial and error the numbers13 and −10 satisfy the condition

use this to split the middle term of

2n² + 3n − 65 = 0

.2n² + 13n − 10n − 65 = 0

n(2n + 13) − 5(2n + 13) = 0

(n − 5)(2n + 13) = 0

solving

n − 5 = 0 ⇒ n = 5

2n + 13 = 0 ⇒ n = −13/2 (rejecct)

required positive integers are

n=5

n+1 =6

n+2=7

for a video of the solution

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   quadratic equation word problem  formation and solution by splitting the middle term

∫ cos x / [(2 + sin x)(4 + sin x)] dx

 find ∫ cos x / [(2 + sin x)(4 + sin x)] dx 

integration by substitution and partial fractions

Substitution

Let t = sin x

Then dt = cos x dx

 ∫ cos x / [(2 + sin x)(4 + sin x)] dx =  ∫ 1 / [(2 + t)(4 + t)] dt

Method of partial fractions

 1 / [(t + 2(t + 4)] = A/(t + 2) + B/(t + 4)

multiply [t+2][t+4]

1 = A(t + 4) + B(t + 2)

Put t = −2

1 = A(2) 

⇒ A = ½

Put t = −4:

1 = B(−2) 

⇒ B = −½

1 / [(t + 2)(t + 4)] = ½[1/(t + 2) − 1/(t + 4)]

integrate

I = ½  ∫ [1/(t + 2)  − 1/(t + 4)] dt 

= ½[ln|t + 2| − ln|t + 4|] + C

= ½ ln|(t + 2)/(t + 4)| + C   use  t = sin x

=½ ln[(2 + sin x)/(4 + sin x)] + C

for  explanation watch this video 

The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

 The sum of numerator and denominator of a fraction is 4 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃. Find the fraction.

Let numerator = x, denominator = y. Fraction = x⁄y

Given

sum of numerator and denominator of a fraction is 4 less than twice the denominator

 x + y = 2y − 4

⇒ x = y − 4  --- (1)

Given

If each of the numerator and denominator is decreased by 1, the fraction becomes ¹⁄₃

 (x − 1)⁄(y − 1) = ¹⁄₃

⇒ 3(x − 1) = y − 1

⇒ 3x − 3 = y − 1

⇒ 3x = y + 2  --- (2)

Solve using substitution method

Substitute (1) into (2):

3(y − 4) = y + 2

3y − 12 = y + 2

2y = 14

y = 7 

 Then 

x = y − 4 

x = 7 -4

x= 3  

x =3 

y=7

 The fraction is ³⁄₇

linear equations in two unknowns

formation of linear equations and solution by substitution method

cbse 10th maths previous years question papers 

2025 2026

for a detailed video of the solution using elimination method

\refer  

If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

 If cos A + sin A = (√2) cos A, prove that cos A − sin A = (√2) sin A

given

cos A + sin A = (√2) cos A

sin A =  (√2) cos A - cos A

sin A = (√2 - 1 ) cos A

sin A/ (√2 - 1 )  = cos A

rationalising

(√2 + 1 ) sin A/ [ (√2 - 1 ) (√2 + 1 ) ]  = cos A

(√2 + 1 ) sin A / [2-1] = cosA

(√2 + 1 ) sin A / [1] = cosA

(√2 + 1 ) sin A = cosA

apply distribution law

(√2) sin A + 1 sinA = cosA

(√2) sin A + sin A = cosA  

(√2) sin A  = cosA - sin A

OR

cos A − sin A = (√2) sin A

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video of an alternate method 

AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the length PA.

 AB is a chord of length 24 cm of a circle of radius 15 cm. The tangents at A and B intersect at a point P. Find the length PA.

Let O be the centre of the circle and Q the midpoint of chord /AB

Given: OA = OB = 15 cm, AB = 24 cm

Q is midpoint of AB

AQ = QB = 24/2 = 12 cm,

also OQ ⊥ AB  

In right triangle  ΔOQA:

OA² = OQ² + AQ²

15² = OQ² + 12²

225 = OQ² + 144

OQ² = 81 

OQ = 9 cm

  ∠OAP = 90° since radius ⊥ tangent

so ΔOAQ ~ ΔOPA

using corresponding sides

OA/OP = OQ/OA

OA² = OQ × OP

15² = 9 × OP

225 = 9 × OP 

 OP = 25 cm 

In right triangle  ΔOPA:

 PA² = OP² − OA²

PA² = 25² − 15² = 625 − 225 = 400

PA = 20 cm  

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circles tangents to circles

a detailed video on an alternative method to solve the problem