Solving Differential Equations Exe-17C Class 12 OP Malhotra ISC Maths Solutions

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Solving Differential Equations Exe-17C Class 12 OP Malhotra ISC Maths Solutions Ch-17 . In this article you would learn about solving a differential equations . Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Solving Differential Equations Exe-17C Class 12 OP Malhotra ISC Maths Solutions

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Differential Equations Class 12 OP Malhotra Exe-17C ISC Maths Solutions Ch-17

Board ISC
Publications  S Chand
Subject Maths
Class 12th
Chapter-17 Differential Equations
Writer OP Malhotra
Exe-17(c) Solving a Differential Equations

Solving a Differential Equations

Differential Equations Class 12 OP Malhotra Exe-17C Solutions

Que-1: dy/dx = 5x + 7

Sol:
Que-1: dy/dx = 5x + 7

Que-2: dy/dx = sin x – x

Sol: Given dydx = sin x – x
⇒ dy = (sin x – x)dx;
On integrating ; we have
∫ dy = ∫ (sin x – x) dx
⇒ y = – cos x – x²/2 + C,
which is the required solution.

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Que-3: dy/dx = x log x

Sol:
Que-3: dy/dx = x log x

Que-4: dy/dx + 2x = e3x

Sol: Given dydx + 2x = e3x
⇒ dy = (e3x – 2x) dx ;
On integrating ; we have
∫ dy = ∫ (e3x – 2x) dx
⇒ y =e3x/3 – x² + C,
which is the required solution.

Que-5: (x + 1)dy/dx = x²

Sol:
Que-5: (x + 1)dy/dx = x²

Que-6: (x + 1)²dy/dx = xex

Sol:
Que-6: (x + 1)²dy/dx = xex

Que-7: dy/dx = sin³ x cos² x + xex

Sol:
Que-7: dy/dx = sin³ x cos² x + xex

Que-8: dy/dx=1/sin4x+cos4

Sol:
Que-8: dy/dx=1/sin4x+cos4x 

Que-9: dy/dx=xsin²x+1/xlogx

Sol:
Que-9: dy/dx=xsin²x+1/xlogx

Que-10: √a+x dy + xdx = 0

Sol: Given diff. eqn. be,
Que-10: √a+x dy + xdx = 0
which is the required solution.

Que-11: dy/dx = √4−y²

Sol:
Que-11: dy/dx = √4−y²

Que-12: dy/dx = sec y

Sol: Given dy/dx = sec y ⇒ dy/dx=1/cosy
⇒ cos y dy = dx
On integrating ; we have
∫ cos y dy = ∫ dx
⇒ sin y = x + c
be the required solution.

Que-13: dy/dx=2-y

Sol: Given dy/dx=2-y
⇒ 1/2-y dy = dx
⇒ 2y dy = dx
On integrating, we have
∫2ydy=∫dx⇒2y/log2=x+c/log2
⇒2y= x log2 + c
which is the required solution.

Que-14: Find the particular solution of edy/dx = x + 1, given that x = 0, y = 3.

Sol: Give diff. eqn. be edy/dx = x + 1
Taking logarithm on both sides, we have
dy/dx = log (x + 1) ⇒ dy = log (x + 1) dx
On integrating ; we have
Que-14: Find the particular solution of edy/dx = x + 1, given that x = 0, y = 3.
given x = 0, y = 3 ∴ from (1) ; we have
3 = log 1 – 0 + c ⇒ c = 3
Thus eqn. (1) becomes :
y = (x+ 1) log (x + 1) – x + 3
be the required solution.

Que-15: Find the particular solution of the differential equation log(dy/dx) = 3x + 4y, given that y = 0 when x = 0.

Sol: Given diff. eqn. be,
Que-15: Find the particular solution of the differential equation log(dy/dx) = 3x + 4y, given that y = 0 when x = 0.
which is the required solution.

–: End of Differential Equations Class 12 OP Malhotra Exe-17C ISC Math Ch-17 Solution :–

Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions

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