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Find solution of the differential equation that satisfies the given initial condition:?
1)
dy/dx = [ycos(x)] / [ 7+y^2 ] ; y(0) = 1
find, 7ln(|y|) + [(y^2) / 2] = ???
2)
xcos(x) = (2y+e^(3y) )y' ; y(0) = 0
find, e^(3y) + 3y^2 = ???
1 Answer
- MathmomLv 71 decade agoFavorite Answer
dy/dx = y cos(x) / (7 + y²)
(7+y²)/y dy = cos(x) dx
(7/y + y) dy = cos(x) dx
Integrate both sides:
∫ (7/y + y) dy = ∫ cos(x) dx
7 ln|y| + y²/2 = sin(x) + C
Use initial condition: y(0) = 1
7 ln 1 + 1/2 = sin(0) + C
0 + 1/2 = 0 + C
C = 1/2
7 ln|y| + y²/2 = sin(x) + 1/2
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x cos(x) = (2y+e^(3y)) dy/dx
x cos(x) dx = (2y + e^(3y)) dy
(e^(3y) + 2y) dy = x cos(x) dx
Integrate both sides:
∫ (e^(3y) + 2y) dy = ∫ x cos(x) dx
1/3 e^(3y) + y² = x sin(x) + cos(x) + C₀
e^(3y) + 3y² = 3x sin(x) + 3cos(x) + C
Use initial condition: y(0) = 0
e^0 + 0 = 0 + 3cos(0) + C
1 = 3 + C
C = -2
e^(3y) + 3y² = 3x sin(x) + 3cos(x) - 2
