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3 years ago

Need help with 6 math problems please

Mathematics

1 answer:

Sav [38]3 years ago

7 0

Answer:

9) 60

10) 16

11) 12

12) 3

13) 60

14) 4

Step-by-step explanation:

9) (2)(5)(6)

(10)(6)

10) 4(5-1)

4(4)

11) 3+4+2+6

7+2+6

9+6

12) 3- (1- $\frac{6}{6}$ )

3- (0)

13) 5(6+1+6-1)

5(7+6-1)

5(13-1)

5(12)

14) 2-3(1-1)+4

2-3(0)+4

-1(0)+4

0+4

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ruslelena [56]

The given ODE

$\underbrace{(y^2+2x)}_M\,\mathrm dx+\underbrace{(2xy-1)}_N\,\mathrm dy=0$

is exact if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ . It so happens that we have $\frac{\partial M}{\partial y}=2y=\frac{\partial N}{\partial x}$ , so it is indeed exact. For such an ODE, we're looking for a solution of the form $\Psi(x,y)=C$ , for which the differential is

$\mathrm d\Psi=\dfrac{\partial\Psi}{\partial x}\,\mathrm dx+\dfrac{\partial\Psi}{\partial y}\,\mathrm dy=0$

so we have the following system of PDEs that allow us to solve for $\Psi$ :

$\dfrac{\partial\Psi}{\partial x}=M=y^2+2x$

$\dfrac{\partial\Psi}{\partial y}=N=2xy-1$

In the first PDE, we can integrate both sides with respect to $x$ and recover $\Psi$ :

$\displaystyle\int\frac{\partial\Psi}{\partial x}\,\mathrm dx=\int(y^2+2x)\,\mathrm dx$

$\implies\Psi(x,y)=xy^2+x^2+f(y)$

Then differentiating this with respect to $y$ returns $N$ :

$\dfrac{\partial\Psi}{\partial y}=2xy+\dfrac{\mathrm df}{\mathrm dy}=2xy-1$

$\implies\dfrac{\mathrm df}{\mathrm dy}=-1$

$\implies f(y)=-y+C$

So the general solution to the ODE is

$\Psi(x,y)=xy^2+x^2-y+C=C$

or simply

$xy^2+x^2-y=C$

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Need help with 6 math problems please ​

Need help with 6 math problems please