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QUESTION: 1

The solution of the differential equation is :

Solution:

dy/dx = y/x - (y^{2}/x^{2} + 1)^{½}…………………(1)

the equation is homogenous,so y = Vx

dy/dx = V + xdv/dx

Put the value of dy/dx in eq(1)

V + xdv/dx = V - (V^{2} + 1)^{½}

= ∫-dv(v^{2} + 1)^{½} = ∫dx/x

= log|v + (v^{2} + 1)^{½}| = -logx + logc

= log|y/x + ((y/x)^{2} + 1)^{½}| = log c/x

=|y/x + ((y/x)^{2} + 1)^{½}| = c/x

QUESTION: 2

The solution of the differential equation is :

Solution:

dy/dx = (x^{2} + 3y^{2})/2xy………….(1)

Let y = vx

dy/dx = v + xdv/dx

Substitute the value of y and dy/dx in (1)

v + x dv/dx = (1+3v^{2})/2v

x dv/dx = (1+3v^{2})/2v - v

x dv/dx = (1 + 3v^{2} - 2v^{2})/2

x dv/dx = (1+ v^{2})/2v

2v/(1+v^{2}) dv = dx/x…………(2)

Integrating both the sides

∫2v/(1+v^{2}) dv = ∫dx/x

Put t = 1 + v^{2}

dt = 2vdv

∫dt/t = ∫dx/x

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

=> log|t/x| = log|c|

t/x = +- c

(1+v^{2})/x = +-c

(1 + (y^{2})/(x^{2}))/x = +-c

x^{2} + y^{2} = Cx^{3}……….(3)

y(1) = 3

1 + 9 = c(1)^{3}

c = 10

From eq(3), we get x^{3}+ y^{2} = 10x^{3}

QUESTION: 3

The first order, first degree differential equation y’ = f(x,y) is said to be homogeneous, if

Solution:

QUESTION: 4

The solution of the differential equation is :

Solution:

Given x dy/dx = y(log y - log x+1)

logy-logx=log(y/x)

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

substituting y=vx

dy/dx=v+xdv/dx

=>dv/(vlogv)=dx/x, integrating we get

log(logv)=log(cx)

log(y/x) = cx

=> y/x = e^{(cx)}

=> y = xe^{(cx)}

QUESTION: 5

The solution of the differential equation is :

Solution:

This is clearly a Homogenous differential equation, as RHS is expressed only in terms of y/x.

To Solve this, Lety/x=t

⟹ y=xt

⟹ dy/dx = t+xdt/dx

By (i)

⟹ tant+t = t+xdt/dx

⟹ tant=xdt/dx

⟹ dx/x=dt/tant

Integrating both sides,

⟹ ∫dx/x = ∫cot tdt

⟹logx = log(sint)+logC

⟹ logx=log(C siny/x)

⟹ x = Csin(y/x)

QUESTION: 6

The solution of the differential equationis :

Solution:

x -(xy)^{½} dy = ydx

[x-(xy)^{½}] - y = dx/dy

dx/dy = x/y - [(xy)^{½}]/y……………….(1)

Let V = x/y

x = Vy

dx/dy = V + ydv/dy……..(2)

V + ydv/dy = V - (V)^{½}

ydv/dy = -(V)^{½}

dv/(V)^{½ }= -dy/y

Integrating both the sides, we get

(V^{(-½+1}))/(-½ + 1) = -log y + c

2(V)^{½} = -log y + c

2(y/x)^{½} = -log y + c

2(x)^{½} = (y)^{½}(-log y + c)

2(x)^{½ }= (y)^{1}/2log y + c(y)^{½}

(x)^{½} = [(y)^{½}]/2 log y + [c(y)^{½}]/2

QUESTION: 7

The solution of the differential equation is :

Solution:

y/x cos y/x dx − (x/y sin y/x + cosy/x) dy = 0

⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x)

This is a homogeneous differential equation.

Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v^{2}cosv)/(sinv+vcosv)

⇒ xdv/dx = (v^{2}cosv)/(sinv+vcosv) - v

⇒x dv/dx = (v^{2} cosv - v^{2} cosv - vsinv )/(sinv+vcosv)

⇒ x dv/dx = - [v sinv/(sinv + vcosv)]

⇒ ∫[(sinv + vcosv)/v sinv]dv = ∫dx/x

Integrating both the sides, we get

∫(cot v + 1/v)dv = - ln(x) + c

ln(sin v) + ln(v) = -ln(x) + c

ln(sin(y/x) + ln(y/x) + ln(x) = c

ln(y/x sin y/x * x) + c

y sin(y/x) = c

QUESTION: 8

The solution of the differential equation x^{2}dy - (x^{2} + 3xy + 4y^{2}) dx = 0 is :

Solution:

QUESTION: 9

The solution of differential equation x^{2}dy + y(x + y)dx = 0 when x = 1, y = 1 is:

Solution:

x^{2}dy + (xy + y^{2}) dx = 0

⇒ x^{2} dy = - (xy + y^{2})dx

⇒ .......(i)

Let y = vx

Differentiating w.r.t x we get

Substituting the value of y and dy/dx in equation (1), we get:

Integrating both sides, we get:

......(ii)

Now, it is given that y = 1 at x = 1.

Substituting D = 1/3 in equation (2), we get

QUESTION: 10

The solution of the differential equation is :

Solution:

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