Can someone give me the answer for this

Mathematics

1 answer:

ella [17]3 years ago

6 0

your answer is areas. area will be scaled by K^2

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Solve the differential equations dy/dx=((xy^2)+x)/y

arsen [322]

$\displaystyle \dfrac{dy}{dx}=\dfrac{xy^2+x}{y}\\ \dfrac{dy}{dx}=\dfrac{x(y^2+1)}{y}\\ \dfrac{y}{y^2+1}\, dy=x\, dx\\ \int \dfrac{y}{y^2+1}\, dy=\int x\, dx\\ \dfrac{\ln (y^2+1)}{2}=\dfrac{x^2}{2}+C\\ \ln(y^2+1)=x^2+C\\ y^2+1=e^{x^2+C}\\ y^2=e^{x^2+C}-1\\ y=\sqrt{e^{x^2+C}-1} \vee y=-\sqrt{e^{x^2+C}-1}\\ \boxed{y=\sqrt{Ce^{x^2}-1} \vee y=-\sqrt{Ce^{x^2}-1}}\\$

7 0

3 years ago

Michaels test scores for this semester were 83,80,91,74,66,99,87,72,95,98. What was the median of his scores

atroni [7]

Answer:

Step-by-step explanation:

Write the numbers in order

66, 72, 74, 80, 83, 87, 91, 95, 98, 99

Find the middle number

83 and 87 are in the middle, so find the mean of them. 83+87= 170

170/2 = 85.

5 0

2 years ago

Read 2 more answers

Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exact<br><br> C=.

Tpy6a [65]

Answer with Step-by-step explanation:

The given differential equation is

$(2x+5y)dx+(5x-4y)dy=0$

Now the above differential equation can be re-written as

$P(x,y)dx+Q(x,y)dy=0$

Checking for exactness we should have

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5$

$\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5$

As we see that the 2 values are equal thus we conclude that the given differential equation is exact

The solution of exact differential equation is given by

$u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)$

The value of $\phi (y)$ can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)

$\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c$