If (t + 1)dx = (2x + (t + 1)3)dt and x(0) = 2 then x(1) is equal to
If (t + 1)dx = (2x + (t + 1)3)dt and x(0) = 2 then x(1) is equal to
Option 1 -
5
Option 2 -
12
Option 3 -
6
Option 4 -
8
-
1 Answer
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Correct Option - 2
Detailed Solution:(t + 1)dx = (2x + (t + 1)3)dt
I.F.
Solution is
x = (t + c) (t + 1)2
x (0) = 2 then c = 2
x = (t + 2) (t + 1)2
x (1) = 12
Similar Questions for you
IF =
So, y(1 + cos2 x) =
y(1 + cos2 x) = – cos x + c
y(0) = 0
0 = – 1 + c
-> c = 1
Now,
so
When x = 0, y = 0 gives
So, for x = 2, y = 12
= cos x – 2 cos2 x =
dy/√ (1-y²) = dx/x²
sin? ¹ (y) = -1/x + c ⇒ c = π/2
sin? ¹ (y) = -π/3 + π/2 = π/6
y = 1/2
. xdy - ydx - x² (xdy + ydx) + 3x? dx = 0
⇒ (xdy - ydx)/x² - (xdy + ydx) + 3x²dx = 0 ⇒ d (y/x) - d (xy) + d (x³) = 0
Integrate both side, we get
y/x - xy + x³ = c
Put x = 3, y = 3
⇒ 1 - 9 + 27 = c
c = 19
Put x = 4
y/4 - 4y = 19 - 64
⇒ y = 12
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