(a) what can you say about a solution of the equation ( y = -(1/6)y^{2} ) just by looking at the…

(a) what can you say about a solution of the equation ( y = -(1/6)y^{2} ) just by looking at the differential equation?\nthe function ( y ) must be strictly increasing on any interval on which it is defined.\nthe function ( y ) must be decreasing (or equal to 0) on any interval on which it is defined.\nthe function ( y ) must be strictly decreasing on any interval on which it is defined.\nthe function ( y ) must be increasing (or equal to 0) on any interval on which it is defined.\nthe function ( y ) must be equal to 0 on any interval on which it is defined.\n(b) verify that all members of the family ( y = 6/(x + c) ) are solutions of the equation in part (a).\n( y=\frac{6}{x + c}rightarrow y=-\frac{square}{(x + c)^{2}} ).\n( lhs=y=-\frac{square}{(x + c)^{2}}=-\frac{1}{6}(\frac{square}{x + c})^{2}=-\frac{1}{6}y^{2}=rhs )\n(c) can you think of a solution of the differential equation ( y = -(1/6)y^{2} ) that is not a member of the family in part (b)?\n( y = 0 ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n( y = e^{6x} ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\nevery solution of ( y = -(1/6)y^{2} ) is a member of the family in part (b).\n( y = 6 ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n( y = x ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n(d) find a solution of the initial - value problem.\n( y = -(1/6)y^{2}quad y(0)=0.1 )\n( y=square )

(a) what can you say about a solution of the equation ( y = -(1/6)y^{2} ) just by looking at the differential equation?\nthe function ( y ) must be strictly increasing on any interval on which it is defined.\nthe function ( y ) must be decreasing (or equal to 0) on any interval on which it is defined.\nthe function ( y ) must be strictly decreasing on any interval on which it is defined.\nthe function ( y ) must be increasing (or equal to 0) on any interval on which it is defined.\nthe function ( y ) must be equal to 0 on any interval on which it is defined.\n(b) verify that all members of the family ( y = 6/(x + c) ) are solutions of the equation in part (a).\n( y=\frac{6}{x + c}rightarrow y=-\frac{square}{(x + c)^{2}} ).\n( lhs=y=-\frac{square}{(x + c)^{2}}=-\frac{1}{6}(\frac{square}{x + c})^{2}=-\frac{1}{6}y^{2}=rhs )\n(c) can you think of a solution of the differential equation ( y = -(1/6)y^{2} ) that is not a member of the family in part (b)?\n( y = 0 ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n( y = e^{6x} ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\nevery solution of ( y = -(1/6)y^{2} ) is a member of the family in part (b).\n( y = 6 ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n( y = x ) is a solution of ( y = -(1/6)y^{2} ) that is not a member of the family in part (b).\n(d) find a solution of the initial - value problem.\n( y = -(1/6)y^{2}quad y(0)=0.1 )\n( y=square )

Answer

(a)

Brief Explanations:

The sign of (y') determines if (y) is increasing or decreasing. Since (y'=-\frac{1}{6}y^{2}), and (y^{2}\geq0) for all real (y), then (y'\leq0). If (y'\lt0), the function is strictly decreasing; if (y' = 0) (when (y = 0)), the function is constant (a special case of non - increasing).

Answer:

The function (y) must be decreasing (or equal to (0)) on any interval on which it is defined.

(b)

Explanation:

Step1: Differentiate (y=\frac{6}{x + C})

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 6), (u^\prime=0), (v=x + C), (v^\prime = 1). Then (y^\prime=\frac{0\times(x + C)-6\times1}{(x + C)^{2}}=-\frac{6}{(x + C)^{2}})

Step2: Substitute into (y') and (y)

(LHS=y^\prime=-\frac{6}{(x + C)^{2}}), and (y=\frac{6}{x + C}). Then (-\frac{1}{6}y^{2}=-\frac{1}{6}(\frac{6}{x + C})^{2}=-\frac{6}{(x + C)^{2}})

Answer:

(y^\prime=-\frac{6}{(x + C)^{2}}), (LHS=y^\prime=-\frac{6}{(x + C)^{2}}), (-\frac{1}{6}(\frac{6}{x + C})^{2})

(c)

Brief Explanations:

For (y = 0), (y^\prime=0) and (-\frac{1}{6}y^{2}=0). The family (y=\frac{6}{x + C}) is not defined when (y = 0) (since solving (0=\frac{6}{x + C}) for (x) gives no real solution). For (y = e^{6x}), (y^\prime=6e^{6x}) and (-\frac{1}{6}y^{2}=-\frac{1}{6}e^{12x}\neq6e^{6x}). For (y = 6), (y^\prime = 0) and (-\frac{1}{6}y^{2}=-6\neq0). For (y=x), (y^\prime = 1) and (-\frac{1}{6}y^{2}=-\frac{1}{6}x^{2}\neq1)

Answer:

(y = 0) is a solution of (y^\prime=-(1/6)y^{2}) that is not a member of the family in part (b).

(d)

Explanation:

Step1: Use the general solution (y=\frac{6}{x + C})

Given (y(0)=0.1=\frac{6}{0 + C})

Step2: Solve for (C)

Cross - multiply: (0.1C=6), so (C = 60)

Answer:

(y=\frac{6}{x + 60})