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2 years ago

9

GIVING 50 POINTS FOR THE FIRST 2 PEOPLE

2 answers:

pochemuha2 years ago

7 0

Answer:Thanks

Step-by-step explanation:

sweet-ann [11.9K]2 years ago

6 0

Answer:

Thank you(:

Step-by-step explanation:

raji~

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(X^2+y^2+x)dx+xydy=0<br> Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

<em>is</em> exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of <em>x</em>, and dividing this by $N(x,y)=xy$ gives an expression independent of <em>y</em>. If we assume $\mu=\mu(x)$ is a function of <em>x</em> alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by <em>x</em> on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution <em>F</em> satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to <em>x</em> gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to <em>y</em> gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

5 0

3 years ago

PLEASE ANSWER THIS ASAP!!!

lora16 [44]

Answer:

so --#9 siunuens sidbishdidhdh did he idbd

4 0

3 years ago

Three Step Linear Equations<br><br> Solve for x<br> 8x+5=-6x+3

tensa zangetsu [6.8K]

Answer:

x = -1/7 or -0.14285 - in decimal form

Step-by-step explanation:

8x + 5 = -6x + 3

8x + 6x = 3 - 5 { change side }

14x = -2 { basic addition and subtraction }

x = -2/14 { changing sides }

x = -1/7 { simplified }

8 0

1 year ago

Read 2 more answers

Which expression is the result of subtracting (Z2-3i)<br> from (Z1 + 2)?<br> A<br> B<br> C<br> D

Reika [66]

Answer:

im not sure but ill try i think A

4 0

2 years ago

Read 2 more answers

Show that line passing through points (3,-2) and (-4,-2) is a horizontal line

Ierofanga [76]

Step-by-step explanation:

horizontal lines will have a gradient(m) of 0

so if m=0 , the line is horizontal

formula to find m:

m=y2-y1 / x2-x1

insert coordinates into formula

$m = \frac{ - 2 - ( - 2)}{ - 4 - 3} \\ \\ m = \frac{ - 2 + 2}{ - 7} \\ m = \frac{0}{ - 7} \\ m = 0$

As m=0 , therefore it is a horizonta line.

hope this helps..

3 0

2 years ago