I\'m having a tough time understanding how to go about answering this question,

Question

I'm having a tough time understanding how to go about answering this question, any help would be appreciated ........................................................................................................................................................................................................................ a) What can you say about a solution of the equation y' = -y^2 just by looking at the differential equation?............................. Answer: that the function is decreasing or equal to 0 ........................................................................................................................ b) Verify that all members of the family y = 1/(x + C) are solutions of the equation in part (a) ..................................................... I'm not too sure how to go about this one. ............................................................................................................................................. c) Can you think of a solution of the differential equation y' = -y^2 that is not a member of the family in part (b)? ................... Because of now really being sure how to answer the previous question, I'm in the same boat here. ...................................... d) Find a solution of the initial-value problem: y' = -y^2, y(0) = 0.5

Explanation / Answer

For (a), revisit our friend from algebra, the parabola. If the sign is positive, it opens upward or to the right. The minimum value is at the vertex. It increases without bound otherwise.
If the sign is negative, it opens downward or to the left. Its maximum value is at the vertex. It decreases without bound otherwise.

For (b), don't be afraid of the constant of integration C. Remember, it's something numeric like four, 59, negative square root of five. The family members differ according to the values of C. If y = 1/(x + C), differentiate it for the left-hand side, square it for the right-hand side and see if you have a true statement.

y' = - y2

y = (x + C)-1

Differentiate using the power rule for u-1, where u = x + C. du = 1 + 0 = 1

y' = - (x + C)-2(1)

y2 = (x + C)-2

Plugging in y'

- (x + C)-2 = - y2

Plugging in y2

- (x + C)-2 = - (x + C)-2

I can't think of any other function either that would solve the differential equation.

For part (d), the initial value lets us determine which family member we're dealing with.

y' = - y2 where y(0) = .5

y(x) = (x + C)-1

y(0) = (0 + C)-1 = .5

1/C = .5

C = 2

y = (x + 2)-1

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