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The differential equation dy + 4y = 5 dx
is valid in the domain 0 ≤ x ≤ 1 with y(0) = 2.25. The solution of the differential equation is
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- y = e4x + 5
- y = e–4x + 1.25
- y = e–4x + 5
- y = e4x + 1.25
- y = e4x + 5
Correct Option: B
+ 4y = 5 | ||
dx |
I.F. = e∫4 dx = e4 x
⇒ y × I.F. = ∫ 5 × (I.F.)dx + C
⇒ y . e4 x = 5. | + C | |
4 |
⇒ 2.25 × 1 = 5. | + C | |
4 |
⇒ C = 1
y . e4 x = 5. | + 1 | |
4 |
⇒ y = | + e-4 x | |
4 |
⇒ y = e-4 x + 1.25