Класс 12 Решения RD Sharma – Глава 22 Дифференциальные уравнения – Упражнение 22.5 | Набор 2

Вопрос 14. sin ⁴ x(dy/dx) = cosx

Решение:

We have,

sin⁴x(dy/dx) = cosx          

dy = (cosx/sin⁴x)dx

Let, sinx = z

On differentiating both sides, we get 

cosx dx = dz

dy = (dz/z⁴)

On integrating both sides, we get

∫(dy) = ∫(1/z⁴)dz

y = (1/ -3t³) + c

y = -(1/3sin³x) + c

y = (-cosec³x/3) + c          -(Here, ‘c’ is integration constant)

Вопрос 15. cosx(dy/dx) – cos2x = cos3x

Решение:

We have,

cosx(dy/dx) – cos2x = cos3x     

(dy/dx) = (cos3x + cos2x)/cosx

(dy/dx) = (4cos³x – 3cosx + 2cos²x – 1)/cosx

(dy/dx) = (4cos³x/cosx) – 3(cosx/cosx) + 2(cos²x/cosx) – secx

dy = [4cos²x – 3 + 2cosx – secx]dx

dy = [4{(cos2x + 1)/2} – 3 + 2cosx – secx]dx

On integrating both sides, we get

∫dy = ∫[2cos2x – 1 + 2cosx – secx]dx

y = sin2x – x + 2sinx – log|secx + tanx| + c          -(Here, ‘c’ is integration constant)

Вопрос 16. √(1 – x ⁴ )(dy/dx) = xdx

Решение:

We have,

 √(1 – x⁴)(dy/dx) = xdx           

Let, x² = z

On differentiating both sides, we get 

2xdx = dz

xdx = (dz/2)

√(1 – z²)dy = (dz/2)

dy = 

On integrating both sides, we get

∫dy = ∫

y = (1/2)sin^-1(z) + c

y = (1/2)sin^-1(x²) + c          -(Here, ‘c’ is integration constant)

Вопрос 17. (а + х)(dy) + хдх = 0

Решение:

We have,

√(a + x)(dy) + xdx = 0       

dy = dx

Let, (x + a) = z²

On differentiating both sides, we get 

dx = 2zdz

(x + a) = z²

x = z² – a

dy = -2[(z² – a)/z]zdz

On integrating both sides, we get

∫dy = -2∫[(z² – a)/z]zdz

y = -(2/3)(z³) + 2az + c

y = -(2/3)(x + a)^3/2 + 2a√(x + a) + c          -(Here, ‘c’ is integration constant)

Вопрос 18. (1 + x ² )(dy/dx) – x = 2tan ^-1 x

Решение:

We have,

(1 + x²)(dy/dx) – x = 2tan^-1x      

(1 + x²)(dy/dx) = 2tan^-1x + x

dy/dx = 

dy 

On integrating both sides, we get

y = I₁ + I₂

I₁ = ∫(

Let, tan^-1x = z

On differentiating both sides, we get 

 = dz

= ∫2zdx

= z²

I₁ = (tan^-1x)²

I₂ = ∫

= (1/2)log|1 + x²|

y = (tan^-1x)² + 1/2log|1 + x²|+ c          -(Here, ‘c’ is integration constant)

Вопрос 19. (dy/dx) = xlogx

Решение:

We have,

(dy/dx) = xlogx             

dy = xlogxdx

On integrating both sides, we get

∫dy = ∫xlogxdx

y = log|x|∫xdx – ∫[∫xdx]dx

y = (x²/2)log|x| – ∫(1/x)(x²/2)dx

y = (x²/2)log|x| – ∫(x/2)dx

y = (x²/2)log|x| – (x²/4) + c          -(Here, ‘c’ is integration constant)

Вопрос 20. (dy/dx) = xe ^x – (5/2) + cos ² x

Решение:

We have,

(dy/dx) = xe^x – (5/2) + cos²x          

dy = (xe^x – (5/2) + cos²x) dx

On integrating both sides, we get

∫dy = ∫xe^x dx – 5/2∫dx + ∫cos²x dx

y = ∫xe^x dx – 5/2∫dx + ∫(1 + cos2x)/2 dx

= ∫xe^x dx – 5/2∫dx + 1/2∫dx + 1/2∫cos2x dx

= ∫xe^x dx – 2∫dx + 1/2∫cos2x dx

= x∫e^x dx – ∫(1∫e^x dx)dx – 2x + sin2x/4 dx

= xe^x – e^x – 2x + 1/4sin2x + c

Вопрос 21. (x ³ + x ² + x + 1)(dy/dx) = 2x ² + x

Решение:

We have,

(x³ + x² + x + 1)(dy/dx) = 2x² + x                  

(dy/dx) = (2x² + x)/(x³ + x² + x + 1)

dy = 

On integrating both sides, we get

∫dy = ∫

Let,

2x² + x = Ax² + A + Bx² + Bx + Cx + C

2x² + x = (A + B)x² + (B + C)x + (A + C)

On comparing the coefficients on both sides,

(A + B) = 2

(B + C) = 1

(A + C) = 0

After solving the equations,

A = (1/2)

B = (3/2)

C = -(1/2)

y = (1/2)∫(dx/(x + 1) + 

y = (1/2)log(x + 1) + (3/4)∫dx – (1/2)∫

y = (1/2)log|x + 1| + (3/4)log|x² + 1| – (1/2)tan^-1x + c          -(Here, ‘c’ is integration constant)

Вопрос 22. sin(dy/dx) = k, y(0) = 1

Решение:

We have,

sin(dy/dx) = k,           

(dy/dx) = sin^-1(k)

dy = sin^-1(k)dx

On integrating both sides, we get

∫dy = sin^-1(k)∫dx

y = xsin^-1k + c              -(1)

Put x = 0, y = 1

1 = 0 + c

1 = c

On putting the value of c in equation(1)

y = xsin^-1k + 1

y – 1 = xsin^-1x

Вопрос 23. e ^(dy/dx) = x + 1, y(0) = 3

Решение:

We have,

e^(dy/dx) = x + 1        

(dy/dx) = log(x + 1)

dy = log(x + 1)dx

On integrating both sides, we get

∫dy = ∫log(x + 1)dx

y = log(x + 1)∫dx – ∫[∫dx]dx

y = xlog(x + 1) – ∫[x/(x + 1)]dx

y = xlog(x + 1) – ∫1 - frac{1}{(x+1)}dx

y = xlog(x + 1) – x + log(x + 1) + c

y = (x + 1)log(x + 1) – x + c         -(1)

Put, y = 3, x = 0 in equation(1)

3 = 0 + c

y = (x + 1)log(x + 1) – x + 3

Вопрос 24. с'(х) = 2 + 0,15х, с(0) = 100

Решение:

We have,

c"(x) = 2 + 0.15x               -(1)

On integrating both sides, we get

∫c"(x)dx = ∫(2 + 0.15x)dx

c(x) = 2x + 0.15(x²/2) + c               -(2)

Put, c(0) = 100, x = 0 in equation(2)

100 = 2(0) + 0 + c

c = 100

c(x) = 2x + 0.15(x²/2) + 100

Вопрос 25. x(dy/dx) + 1 = 0, y(-1) = 0

Решение:

We have,

x(dy/dx) + 1 = 0                

xdy = -dx

dy = -(dx/x)

On integrating both sides, we get

∫dy = -∫(dx/x)

y = -logx + c           -(1)

Put, y = 0, x = -1 in equation(1)

0 = 0 + c

c = 0

y = -log|x|

Вопрос 26. x(x ² – 1)(dy/dx) = 1, y(2) = 0

Решение:

We have,

x(x² – 1)(dy/dx) = 1             -(1)     

dy = dx/x(x + 1)(x – 1)

On integrating both sides, we get

∫dy = ∫dx/x(x + 1)(x – 1)

Let, 1/x(x + 1)(x – 1) = A/x + B/(x + 1) + C/(x – 1)

1 = A(x + 1)(x – 1) + B(x)(x – 1) + C(x)(x + 1)             -(2) 

Put, x = 0, -1, 1 respectively and simplify above equation, we get,

A = -1, B = (1/2), C = (1/2)

y = 

y = -logx + (1/2)log(x + 1) + (1/2)log(x – 1)

y = (1/2)log(1/x²) + (1/2)log(x + 1) + (1/2)log(x – 1) + c             -(3) 

Put, y = 0, x = 2 in equation(3)

0 = (1/2)log(1/4) + (1/2)log(3) + 0 + c

c = -(1/2)log(3/4)

y = (1/2)log[(x² – 1)/x²] – (1/2)log(3/4)