Ordinary Differential Equations

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Ordinary Differential Equations
Ordinary Differential Equations

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Ordinary Differential Equations
1.(a). Given Differential equation : dy
dx = y2 + 3y – 4
Substituting y(x) = −v' ( x)
v (x) ,
And also, differentiating y(x) with respect to x-
dy
dx = y’(x) = −v v'' +¿ ¿,
Put it in differential equation : −v v'' +¿ ¿ = (- v'
v )2 + 3(- v'
v ) – 4
-vv’’ + (v’)2 = (v’)2 - 3v’v -4v2
-vv’’ = -3v’v – 4v2
v’’ – 3v’ – 4v = 0 ...(1)
equation (1) is the desired Second Order ODE.
Now, from the differential equation, we have, dy
y2+3 y −4 = dx
dy
( y+4 ) ( y−1) = dx
dy
5 ( y + 4 ) − dy
5( y −1) = dx
Integrating both sides-
ln(5(y + 4))1/5 - ln(5(y – 1))1/5 = x + C (Using the property ln(xy) = yln(x) )
Here, C is the constant of integration. So,
y + 4
y−1 = e5(x + C) (Using the property ln(x) – ln(y) = ln
(x/y) )
Applying Componendo and Dividendo Rule-
2 y +3
5 = e5(x +C)+1
e5 (x+C )−1
Taking e5/ 2( x+C) common from Numerator and denominator-

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Ordinary Differential Equations
2 y +3
5 = e5/ 2(x +C)+e−5 /2( x+C)
e5 / 2(x+C )−e−5 /2 (x+C)
2 y +3
5 = coth(5/2(x + C))
y = ½( 5(coth(5/2(x + C))) – 3) is the required non constant solution.
Also,
Solving the above equation (1) by putting v’’ = D2 operator and v’= D, then, the
auxiliary equation will be-
(D2 – 3D – 4)v = 0
Solving this quadratic equation,
D = 3± √9+16
2 = 3± 5
2
D = 4, -1
Therefore, v(x) of the above second order ODE (eq.1) is ,
v(x) = Ae4x + Be-x
v’(x) = 4Ae4x –Be-x
Therefore,
y(x) = - 4 A e4 x−Be− x
Ae4 x + Be− x is the required two constant solution.
1.(b). Given Cauchy-Euler Equation : 2x2 d2 y
d x2 + 7x dy
dx - 18y = 0
Assuming the solution of this equation to be y(x) = xr only for x > 0, then,
We will get the equation in terms of r as-
2r(r – 1) + 7r – 18 = 0
Solving it-
2r2 + 5r – 18 = 0