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Mathematics, 10.04.2020 21:52 ColungaRosemary

Ordinary Differential Equations
If possible please show me in detail

Ordinary Differential Equations If possible please show me in detail

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Answers: 2

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Answer from: AquaTyrant5054

7.15

$\dfrac y{x^2}\,\mathrm dx-\dfrac{xy+1}x\,\mathrm dy=0$

This is a Bernoulli equation in disguise. Solve for $\frac{\mathrm dx}{\mathrm dy}=x'$ :

$x'=\dfrac{x(xy+1)}y=x^2+\dfrac xy$

Divide through both sides by x^2 :

$x^{-2}x'=1+\dfrac1{xy}$

Substitute $v=x^{-1}$ and $v'=-x^{-2}x'$ , and the resulting ODE is linear in .

$-v'=1+\dfrac vy\implies v'+\dfrac vy=-1$

Multiply both sides by , integrate both sides, solve for , then for :

$yv'+v=(yv)'=-y\implies yv=-\dfrac{y^2}2+C\implies \boxed{x(y)=\dfrac{2y}{C-y^2}}$

8.15

$yy'=-\dfrac x2$

This is separable; write $y'=\frac{\mathrm dy}{\mathrm dx}$ , then

$y\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac x2\implies y\,\mathrm dy=-\dfrac x2\,\mathrm dx$

Integrate both sides to get

$\dfrac{y^2}2=-x^2+C$

Given that y=2 when x=4 , we find

$\dfrac{2^2}2=-4^2+C\implies C=18$

so the particular solution is

$\boxed{\dfrac{y(x)^2}2=18-x^2}$

10.15

The image is a bit blurry, but it looks like the equation is

$xy'''-y''+\dfrac1x=0$

Substitute v=y'' , so that v'=y''' and the equation is linear in :

$xv'-v=-\dfrac1x$

Divide through both sides by x^2 :

$\dfrac{v'}x-\dfrac v{x^2}=\left(\dfrac vx\right)'=-\dfrac1{x^3}$

Integrate both sides and solve for :

$\dfrac vx=\dfrac1{2x^2}+C$

$v=\dfrac1{2x}+Cx$

Solve for by integrating both sides twice:

$y''=\dfrac1{2x}+Cx$

$y'=\dfrac{\ln x}2+C_1x^2+C_2$

$\boxed{y(x)=\dfrac{x(\ln x-1)}2+C_1x^3+C_2x+C_3}$

11.15

y''y^3+25=0

Unfortunately, I'm not sure how to tackle this one just yet...

12.15

y''-y'=3x^2-2x+1

You could use undetermined coefficients here, but since the equation is free of , you can substitute v=y' and v'=y'' to reduce the order and get a linear equation.

v'-v=3x^2-2x+1

Multiply through both sides by $e^{-x}$ , integrate, solve ... you know the drill:

$e^{-x}v'-e^{-x}v=(e^{-x}v)'=(3x^2-2x+1)e^{-x}$

$e^{-x}v=-(3x^2+4x+5)e^{-x}+C$

v=-(3x^2+4x+5)+Ce^x

$\boxed{y(x)=-(x^3+2x^2+5x)+C_1e^x+C_2}$

13.15

y'''+4y''+4y'=(9x+15)e^x

We could reduce the order again, but that won't avoid having to solve a higher-order ODE like in the previous examples. The homogeneous equation

y'''+4y''+4y'=0

has characteristic equation

r^3+4r^2+4r=r(r^2+4r+4)=r(r+2)^2=0

with roots r=0 and r=-2 (with multiplicity 2), and thus the characteristic solution is

$y_c=C_1+C_2e^{-2x}+C_3xe^{-2x}$

For the particular solution, assume an ansatz

y_p=(a_0+a_1x+a_2x^2)e^x

with derivatives

${y_p}'=(a_0+a_1+(a_1+2a_2)x+a_2x^2)e^x$

${y_p}''=(a_0+2a_1+2a_2+(a_1+4a_2)x+a_2x^2)e^x$

${y_p}'''=(a_0+3a_1+6a_2+(a_1+6a_2)x+a_2x^2)e^x$

Substitute these into the ODE and simplify to end up with

(9a_0+15a_1+14a_2+(9a_1+30a_2)x+9a_2x^2)e^x=(9x+15)e^x

Matching up coefficients tells us

$9a_2=0\implies a_2=0$

$9a_1+30a_2=9\implies a_1=1$

$9a_0+15a_1+14a_2=15\implies a_0=0$

So the particular solution is

y_p=xe^x

and the general solution to the ODE is

$\boxed{y(x)=C_1+C_2e^{-2x}+C_3xe^{-2x}+xe^x}$

14.15

$y''+y=2\cos5x+3\sin5x$

The characteristic equation is

r^2+1=0

with roots $r=\pm i$ , which admits the characteristic solution

$y_c=C_1\cos x+C_2\sin x$

Assume the ansatz

$y_p=a\cos5x+b\sin5x$

with second derivative

${y_p}''=-25a\cos5x-25b\sin5x$

Substitute these into the ODE, simplify, and solve for a,b :

$-24a\cos5x-24b\sin5x=2\cos5x+3\sin5x$

$-24a=2\implies a=-\dfrac1{12}$

$-24b=3\implies b=-\dfrac18$

Then the general solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac{\cos5x}{12}-\dfrac{b\sin5x}8}$

ansver

Answer from: Quest

the length of o'p' is $1.5\times op$ .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

$k=\frac{o'p'}{op}$

the scale factor is 1.5.

$1.5=\frac{o'p'}{op}$

multiply both sides by op.

$1.5\times op=o'p'$

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is $1.5\times op$ .

ansver

Answer from: Quest

9 x 8 divided by (5-3)=36

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