Page 104 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 5: Differential Equations

Notes
y
d x 0 ...(iii)
x
y
so that x = constant =a (say a) ...(iv)
x
so the solution of (i) is equation (iv)
(D) Linear Equation of the first order (non-homogeneous)
Let the differential equation of the first order has the form

dy
f x y f x ...(i)
1
2
dx
In order to solve (i), put
y u x v x ...(ii)

dy dv du
Then u v
dx dx dx
So equation (i) becomes
dv du
u v f x u x v x f x
1
2
dx dx
du dv
or v f x u x u f x ...(iii)
2
1
dx dx
we choose u such that
du
f x u x 0
1
dx
du
or f x dx 0 ...(iv)
1
u
Solving (iv) we have
logu f x dx a ...(v)
1
The simplest solution is when a = 0, so that

1 f x dx ...(vi)
u e
From (iii), (iv) and (vi) we have

1 f x dx dv
e f x
2
dx
dv 1 f x dx
or f x e
2
dx
Thus v e 1 f x dx f x dx a 2 ...(vii)
2
1 f x dx 1 f x dx 1 f x dx
so y uv a e e e f x dx ...(viii)
2
2

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