Question
The differential equation dy
is valid in
the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The
solution of the differential equation is
Options :
y = e-4x + 1.25
y = e4x + 5
y = e4x + 1.25
y = e-4x + 5
Answer :
y = e-4x + 1.25
Solution :
P(x) = 4, Q(x) = 5
I.F = e∫P x dx =e∫4dx = e4x
ye4x = ∫5e4xdx+c
ye4x = (5/4)e4x+c
y= (5/4)+ce-4x
Substitute given initial condition, x = 0, y = 2.25
We get, c = 1 and y = 5/ 4 + e-4x
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