Question:

Answer:

Question 1: Find the explicit general solution of xdx – (y^2)dy=0.

Xdx – (y^2)dy = 0

Or, xdx = (y^2)dy

integrating both sides we get,

(x^2)/2 + C1 = (y^3)/3 + C2

Or, 3(x^2) + 3C1 – 2(y^3) – 2C2 = 0

Let, 3C1 – 2C2 = C’

Therefore,

3(x^2) – 2(y^3) + C’ = 0

Hence, this is the general solution of the equation.

Question 2: Find the explicit general solution of 2xyy’ – y^2 +x^2 = 0 (Homogeneous Substitution Method)

2xyy’ – y^2 + x^2 = 0

Or, dy/dx = (x^2 – y^2)

By the substitution method,

Let v = y/x

Therefore, y = vx

Or, dy/dx = v + x(dv/dx)

Therefore, by substituting the value we get,

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

Or, dx/x = 2dv/((3v^2)-1)

Integrating both sides we get,

Logx + C1 = 1/3log ((3v^2)-1) + C2

Or, x^3= 3v^2 – 1 + C

Where C is the composite constant

Putting the value of v in the equation, we get,

x^3= {3(y^2)/(x^2)} – 1 + C

Question 3: Find the explicit general solution of xy’ + y + 4 =0.

a)      Integrating Factor

xdy/dx + y + 4 = 0

Or, dy/dx + y/x = -4/x

By integrating factor method we get,

P = 1/x

Therefore the integrating factor = e^∫dx/x = e^log x = x

Therefore we are required to multiply on both the sides by the integrating factor, we get

(dy/dx + y/x)xdx = -(4/x)xdx

xy = -4x +c

y = c/x – 4“

b) Exact Equation Method

(y+4) + (x)y’= 0

Now by exact equation method we get

My(x,y) = 1

Nx(x,y) = 1

Fx(x,y) = y + 4

Fy(x,y) = x

Fx’(x,y) = y + h(y)

Fx”(x,y) = h’(y)

h’(y) = 0

h(y) = 0

y + 4 = c1/x

Question 4: Find the explicit general solution of y’ – 3y/x= (x^4) y^(1/3) using Bernoulli’s theorem

Let, a(x) = (-3/x)

f(x) = x^4

n=1/3

z= y^(1-n)

Therefore according to the Bernoulli’s theorem,

dz/dx=(1-n) a(x)y^(1-n) + (1-n)f(x)

or,  dz/dx=(1-n) a(x)z + (1-n)f(x)

or, dz/dx = 2/3x^4 – 2z/x

or, dz/dx = (2/3)x^4 – 2z/x

Now approaching the problem with the integrating factor

The integrating factor = x^2

(dz/dx + 2z/x) x^2 = (2x^6)/3

Integrating both sides, we get

(x^2)z = (x^7)/21 + c

(x^2)(y^2/3) = (x^7)/21 + c

Question 5: Find the explicit general solution of y“= -2y` + 4t

Solution to question a

y” + 2y’ = 4t

let y’’ = a^2

therefore

a^2 + 2a = 0

0r a+2 = +-2

Therefore y(t) = c1 e^ -2t +c2 + t^2 + t.

Solution to question b

For y” = 2 and y = 1 and t =0,

c1 + c2 = 1.

Question 6: Find the explicit general solution of (x + y) y’=4

(x + y) y’=4

Let x + y = v

Therefore y= v-x

dy/dx = dv/dx – 1

therefore,

vdv/dx – v = 4

(4+v)/v = dv/dx

dx = vdv/(4+v)

x= v – ln (|v+4|^4) + c

Now putting the value of v we get,

x= x + y – ln (|x+y+4|^4) +c

y = ln (|x+y+4|^4) +c

Question 7: Using Euler’s method for the approximate points on the solution curve; y’= -x/y

Starting point = (0, 1)

h = 0.1

x0 = 0, y0 = 1

x1 = x0 + h = 0.1

x2 = x1 + h = 0.2

x3 = x2 + h = 0.3

x4 = x3 + h = 0.4

y1 = y0 + h f(x0,y0) = 1

y2 = y0 + h f(x1,y1) = .99

y3 = y0 + h f(x2,y2) = .98

y4 = y0 + h f(x3,y3) = .97

Question 8: Find the explicit solution of y’ – 5y = 0

y(0) = 2

L { y’ – 5y = 0}

S Y(S) – 2 + 5 Y(S) = 0

Or, (S+5) Y(S) = 2

Or, Y(S) = 2/(S+5)

L^-1{ Y(S) = 2/(S+5)}

Or, Y(t)= 2e^(−5t)

Question 8: Find the explicit solution of y” – 4y = 0

y (0)= 2

y’ (0)= 2

L { y” – 4y = 0}

Or, (S^2) Y(S) – S.2 – 2 + 4Y(S) = 0

Or, ((S^2) + 4) Y(S) = 2S+2

Y(S) = (2S + 2)/ ((S^2) + 4)

L^-1{ Y(S) = (2S + 2)/ ((S^2) + 4)}

Or, Y(t) = sin2t + 2cos2t