Differential Equations Class 12 OP Malhotra Exe-17D Maths Solutions. In this article you would learn about equation reducible to variable separable. 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.
Differential Equations Class 12 OP Malhotra Exe-17D Maths Solutions Ch-17
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-17 | Differential Equations |
| Writer | OP Malhotra |
| Exe-17(d) | Equation Reducible to Variable Separable. |
Equation Reducible to Variable Separable
Differential Equations Class 12 OP Malhotra Exe-17D Solutions
Solve the following differential equations.
Que-1: (i) dy/dx = ex+y
(ii) dy/dx = x³ e-2y
Sol: (i) Given dy/dx = ex+y = ex ey
⇒ e-y dy = ex dx
[after variable separation]
On integrating ; we have
∫ e-y dy = ∫ ex dx + c
⇒ – e-y = ex + c ⇒ ex + e-y = c’
which is the required solution.
(ii) Given dy/dx = x³ e-2y
Que-2: dy/dx=xy+y/xy+x
Sol:
Que-3: y (1 – x²) dy = x(1 + y²) dx
Sol: Given y (1 – x²) dy = x (1 + y²) dx
Que-4: dy/dx + cosxsiny/cosy = 0
Sol: Given dy/dx + cosxsiny/cosy = 0 ;
after variation separation, we have
cosy/siny+ cos x dx = 0 ; On integrating
⇒ ∫ cot y dy + ∫ cos xdx = c
⇒ log sin y + sin x = c
⇒ log sin y = c – sin x
⇒ sin y = Ae-sinx
which is the required solution.
[where A = ec]
Que-5: (i) (x + xy) dx + (x – xy²) dy = 0
(ii) (x² – yx²) dy + (y²+ xy²) dx = 0
(iii) x² (y + 1) dx + y² (x- 1) dy = 0
Sol: (i) (y + xy) dx + (x – xy²) dy = 0
⇒ y (1 + x) dx + x (1 – y²) dy = 0
⇒ (1+x/x)dx + (1−y²/y)dy = 0
On integrating ; we have
∫(1/x+1) dx+∫1/y dy−∫y dy = c
⇒ log |x| + x + log |y | – y²/2 = c
be the required solution
(ii) (x² – yx²) dy + (y² + xy²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x) dx = 0
⇒ 1−y/y² dy+1+x/x² = 0
On integrating ; we have
∫(1/y²−1/y)dy+∫1/x²dx+∫1/x dx = 0
⇒ – 1y−log|y|−1x+log|x|
⇒ log x/y = x+y/xy + c
be the required solution.
(iii) Given x² (y + 1) dx + y² (x – 1) dy = 0
be the required solution.
Que-6: √(a+x) dy/dx = -xy
Sol: Given√(a+x) dy/dx = -xy; after variable separation, we have
which is the required solution.
Que-7: dy/dx = 1 – x + y – xy
Sol: Given dy/dx = 1 – x + y – xy
= 1 – x + y(1 – x)
⇒ dy/dx = (1 – x)(1 + y)
After variable separation, we have dy/1+y = (1 – x) dx ; on integrating
log | 1 + y | = x – x²/2 + c which is the required solution.
Que-8: (1 + x) (1 + y²) dx + (1 +.y) (1 + x²) dy = 0
Sol: Given (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0
Que-9: sec² x tany dx – sec²y tan x dy = 0
Sol: Given sec² x tan y dx – sec² y tan x dy = 0
On dividing throughout by tan x tan y; we get
sec²xdxtanx−sec²ydytany = 0 ; on integrating
∫sec2xdxtanx−∫sec2ydytany = log c
⇒ log | tan x | – log | tan y | = log c
⇒ log ∣∣tanxtany∣∣ = log c
⇒ tan x = A tan y where A = ± c which is the required solution.
Que-10: dy/dx = ex-y + e2logx-y
Sol:
Que-11: (i) cos x cos y dy + sin x sin y dx = 0
(ii) (1 + cos x) dy = ( 1 – cos y) dx
Sol: (i) Given cos x cos y dy + sin x sin y = 0
after variable separation, we have
cosy/siny dy +sinx/cosx dx = 0; On integrating
= log |sin y| – log |cos x| = log c
⇒ log ∣siny/cosx∣ = log c
⇒ sin y = ± c cos x
⇒ sin y = A cos x
which is the required solution.
(ii) Given (1 + cos x)dy = (1 – cos y) dx
after variable separation, we have
Que-12: (1 + x²)dy + x√(1−y²) dx = 0
Sol:
Que-13: (1 – x²) dy + xy dx = xy² dx
Sol:
Que-14: x√(1+y²) dx+y√(1+x²) dy = 0
Sol: Given diff. eqn. be, x√(1+y²) dx+y√(1+x²) dy = 0
dividing throughout by √(1+x²) √(1+y²); we have
which is the required solution.
Que-15: (ex + 1)y dy = (y + 1)exdx
Sol: Given (ex + 1 )y dy = (y + 1)ex dx; after variable separation, we have
which is the required solution.
Que-16: (i) log(dy/dx)= ax + by
(ii) log(dy/dx) = 3x – 5y
Sol:
Que-17: dy/dx = cos³ x sin4 x + x√(2x+1)
Sol: Given dy/dx = cos³ x sin4 x + x√(2x+1) after variable separation, we have
Que-18: √(1+x²+y²+x²y²) + xy dy/dx = 0
Sol: Given diff. eqn. be, √(1+x²+y²+x²y²) + xy dy/dx = 0
Que-19: dy/dx = ex(sin²x+sin2x)/y(2logy+1)
Sol: Given, dy/dx = ex(sin²x+sin2x)/y(2logy+1) after variable separation, we have
Que-20: √(1+x²) dy + √(1+y²) dx = 0
Sol:
Que-21: dy/dx−xsin²x = 1/xlogx
Sol:
Que-22: Find the particular solution of the following differential equations :
(i) cos y dy + cos x sin y dx = 0, y(π/2) = π/2
(ii) (1 – x²)dy/dx – xy = x; given y = 1 when x = 0.
(iii) (x² – yx²)dy + (y² + x²y²)dx = 0, given that y = 1, when x = 1.
(iv) dy/dx = 1 + x² + y² + x²y², given that y = 1 when x = 0.
(v) (1 + e2x) dy + (1 + y²)exdx = 0, given that y = 1 when x = 0.
(vi) dy/dx = 1 + x + y + xy, given that y = 1 when x = 0.
(vii) dy/dx = x(2log|x|+1)/siny+ycosy, given that y = π/2, when x = 1.
Sol: (i) Given cos y dy + cos x sin y dx = 0 after variable separation, we have
(ii) Given (1 – x²)dy/dx – xy = x ⇒ (1 – x²)dy/dx = x (y + 1)
(iii) Given diff. eqn. be, (x² – yx²) dy + (y² + x²y²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x²) dx = 0 ; after variable separation, we have
(iv) Given diff. eqn. be, dy/dx = 1 + x² + y² + x²y² = (1 + x²)(1 + y²);
after variable separation; we have
(v) Given diff. eqn. be, (1 + e2x) dy + (1 + y²)ex dx = 0 ; after variable separation, we have
(vi) Given dy/dx = 1 + x + y + xy = (1 + x) (1 + y) ⇒ dy1+y = (1 + x)dx;
Que-23: dy/dx = (2x + 3y – 4)²
Sol: Given dy/dx = (2x + 3y – 4)²
put 2x + 3y – 4 = t
Diff. both sides w.r.t. x
which is the required solution.
Que-24: (x + y + 1) dy/dx = 1
Sol: Given diff. eqn. be
where c’ = c – 1
Que-25: (x + y)² dy/dx = 1
Sol: Given diff. eqn. be,
which the required solution.
Que-26: dy/dx = cos(x + y)
Sol:
Que-27: dy/dx = tan²(x + y)
Sol: Given diff. eqn. be,
where c’ = 4c
which is the required solution.
Que-28: cos²(x – 2y) = 1 – 2dy/dx
Sol: Given differential eqn. be,
which is the required solution.
Que-29: dy/dx = x+y+1/2x+2y+3
Sol: Given differential eqn. be,
dydx = x+y+1/2x+2y+3 … (1)
put x + y = t; diff. both sides w.r.t. x ; we get
which is the required solution.
–: End of Differential Equations Class 12 OP Malhotra Exe-17D ISC Math Ch-17 Solution :–
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