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stealth61 [152]

3 years ago

7

Use the zero product property to solve the equation. (4k+5)(k+7)=0

1 answer:

Maru [420]3 years ago

7 0

The zero product property tells us that if we have
xy=0, then we can assume that x and y both equal 0

so

(4k+5)(k+7)=0
we can assume that 4k+5=0 and k+7=0
so

4k+5=0
minus 5 both sides
4k=-5
divide both sides by 4
k=-5/4

k+7=0
minus 7 both sides
k=-7

k=-5/4 or -7

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icang [17]

Answer:

Linear and non-homogeneous.

Step-by-step explanation:

We are given that

$\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}$

We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).

We have also find type of differential equation.

$y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}$

$y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}$

$y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}$

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y'+P(x)y=g(x)

Compare it with first order first degree linear differential equation

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$P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}$

$\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}$

Homogeneous equation

$\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}$

Degree of f and g are same.

$f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x$

Degree of f and g are not same .

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