3 years ago

Compare the rates of change of the following items.

Mathematics

1 answer:

musickatia [10]3 years ago

7 0

The answer is B. The slope in item 2 is steap and steady

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3 years ago

Solve the Differential equation (x^2 + y^2) dx + (x^2 - xy) dy = 0

natita [175]

Answer:

$\frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

Step-by-step explanation:

Given differential equation,

$(x^2 + y^2) dx + (x^2 - xy) dy = 0$

$\implies \frac{dy}{dx}=-\frac{x^2 + y^2}{x^2 - xy}----(1)$

Let y = vx

Differentiating with respect to x,

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

From equation (1),

$v+x\frac{dv}{dx}=-\frac{x^2 + (vx)^2}{x^2 - x(vx)}$

$v+x\frac{dv}{dx}=-\frac{x^2 + v^2x^2}{x^2 - vx^2}$

$v+x\frac{dv}{dx}=-\frac{1 + v^2}{1 - v}$

$v+x\frac{dv}{dx}=\frac{1 + v^2}{v-1}$

$x\frac{dv}{dx}=\frac{1 + v^2}{v-1}-v$

$x\frac{dv}{dx}=\frac{1 + v^2-v^2+v}{v-1}$

$x\frac{dv}{dx}=\frac{v+1}{v-1}$

$\frac{v-1}{v+1}dv=\frac{1}{x}dx$

Integrating both sides,

$\int{\frac{v-1}{v+1}}dv=\int{\frac{1}{x}}dx$

$\int{\frac{v-1+1-1}{v+1}}dv=lnx + C$

$\int{1-\frac{2}{v+1}}dv=lnx + C$

$v-2ln(v+1)=lnx+C$

Now, y = vx ⇒ v = y/x

$\implies \frac{y}{x}-2ln(\frac{y}{x}+1)=lnx+C$

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3 years ago

Isaac walks 6/10 of a mile in 1/5 of an hour. If Isaac’s walking rate remains constant, what is Isaac’s walking rate in miles pe

navik [9.2K]

Answer:

Step-by-step explanation:

u have to mutypl and div it up sum and k/=+56+8=90/h=+5689-097/p=11+y=1+11=12

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