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Differential Equations (MA-101)?
Hello, I've been having some trouble on Differential Equations in Calc class. Any help is appreciated for any of the questions down below (answers, hints, explanation).
Find the solution of the differential equation that satisfies the given initial condition.
1) 6yy' = x, y(6) = 3
2) dx/dt = 3/x, x(0) = 5
3) dy/dt = 8y^5, y(0) = 1
4 dy/dx +y/4 = 0, y(0) = 14
5) [(x^2)/(y^2-7)](dy/dx) = 1/2y, y(1) = sqrt(8)
6) y'(x) = sqrt[-2y(x)+17], y(5) = 4
7) du/dt = e^(2.1t-2.2u), u(0) = 4
Once again, any and all help is greatly appreciated!
4 Answers
- ?Lv 73 years agoFavorite Answer
All these are separable.
So rewrite in the form: f(y) dy = g(x) dx, integrate both sides, then solve for y
5)
I'll assume right side is 1/(2y)
x²/(y²−7) dy/dx = 1/(2y), y(1) = √8
2y/(y²−7) dy = 1/x² dx
∫ y/(y²−7) dy = ∫ 1/x² dx
ln|y²−7| = −1/x + C₀
y² − 7 = e^(C₀−1/x) = e^C₀ * e^(−1/x)
y² − 7 = C e^(−1/x)
y² = C e^(−1/x) + 7
y(1) = √8
(√8)² = Ce^(−1) + 7
8 = C/e + 7
C = e
y² = e * e^(−1/x) + 7
y² = e^(1−1/x) + 7
y = √(e^(1−1/x) + 7)
- alexLv 73 years ago
Hint:
Separation of Variables
7) du/dt = e^(2.1t-2.2u)
---> e^(2.2u)du =e^(2.1t)dt
integrate both sides
...
- ?Lv 73 years ago
First, it considered quite impolite to ask many questions in one post. You should have known or deduced that! Ask one per post.
hints: These are ALL separable. So, rewrite them separated [thats highschool algebra]. Then just integrate!
The constants are then found via the given conditions [that step too is highschool algebra]. Done!
