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QUESTION: 1

Write the order and degree of the given differential equation:

Solution:

Order = 3 and Degree = 1

The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

QUESTION: 2

The order of a differential equation representing a family of curves is same as:

Solution:

The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.

QUESTION: 3

The solution of the equation x^{3}dx + (y + 1)^{2} dy = 0

Solution:

x^{3}dx + (y + 1)dy = 0

=> d(x^{4}/4) + d((y + 1)^{3}/3) = 0

=> d(x^{4}/4 + ((y + 1)^{3}/3)) = 0

=> (x^{4}/4 + ((y + 1)^{3}/3)) = constt

=> 12 *(x^{4}/4 + ((y + 1)^{3}/3)) = 12 * constt

=> 3x^{4} + 4(y + 1)^{3} = constt

QUESTION: 4

Identify the type of differential equation

Solution:

(x−y)dy/dx = x + 2y

⇒ dy/dx = (x + 2y)/(x−y)

F (x,y) = (x + 2y)/(x−y)

F(Ax, Ay) = (Ax + 2Ay)/(Ax−Ay)

= A(x + 2y)/A(x−y)

= (x + 2y)/(x − y) = F (x,y)

Hence, the equation is homogenous.

QUESTION: 5

Identify the form of the given Differential Equation

Solution:

QUESTION: 6

The solution of the initial value problem is :

Solution:

xdy/dx = -coty

dy/dx = - coty/x

dy/coty = - dx/x

tany dy = - dx/x

∫(siny/cosy)dy = - ∫dx/x…………..(1)

Put t = cosy

dt = -sinydy

Put the value of dt in eq(1)

-∫dt/t = -∫dx/x

log |t| = log |x| + log c

log |cosy| = log(c.|x|)

|cos y| = c.|x|....................(2)

As y = π/4, x = (2)^{½}

cos π/4 = c*(2)^{½}

1/(2)^{½} = c*(2)^{½}

c = ½

Put the value of ‘c’ in eq(2)

cos y = x/2 => x = 2cosy

QUESTION: 7

The solution of the differential equation is :

Solution:

dy/dx = e^{ax} cos by

∫dy/cos by = ∫e^{ax} dx

∫sec by dy = ∫e^{ax} dx

= (log| sec by + tan by|)/b = e^{ax} /a + c

= a(log| sec by + tan by|) = be^{ax} + c

QUESTION: 8

The solution of the differential equation is :

Solution:

dy/dx = x logx

=> ∫dy = ∫x logx dx

y = logx . x^{2}/2 - ∫1/x . x^{2}/2 dx + c

y = x^{2}/2 log x -½ ∫x dx + c

y = x^{2}/2 log x - ½ . x^{2}/2 + c

y = x^{2}/2 log x - x^{2}/4 + c

QUESTION: 9

The solution of the differential equation is :

Solution:

d^{2}y/dx^{2} = xe^{x}

d^{2}y/dx^{2} = ∫xe^{x}

d/dx(dy/dx) = [xe^{x}]

Integrating, we get

dy/dx = (x-1)e^{x}+ c1

Again integrating, y = (x-2)e^{x} + c1x + c2

QUESTION: 10

The solution of the initial value problem e^{dy/dx} = x + 1, y(0) = 3 is :

Solution:

e^{dy/dx} = x + 1

⇒ dy/dx = log(x+1)

⇒ dy = log(x+1)dx

Integrating both sides, we get

⇒ ∫dy = ∫log(x+1)dx

y = log(x + 1) ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx

y = xlog(x + 1) - ∫1/(x+1)xdx

y = xlog(x + 1) - ∫(1 - 1/(x + 1)dx

y = xlog(x + 1) - ∫dx + ∫1/(x+1)dx

y = xlog(x + 1) - x + log|x + 1| + c

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

Here c is 3

y = (x + 1)log|x + 1| - x + 3

### Separable differential equations 2

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