Ordinary Differential Equations If possible please show me in detail - QAmmunity.org

Ordinary Differential Equations If possible please show me in detail

asked Sep 21, 2021 118k views

1 Answer

7.15

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

This is a Bernoulli equation in disguise. Solve for
$(\mathrm dx)/(\mathrm dy)=x'$ :

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

Divide through both sides by
x^2 :

$x^(-2)x'=1+\frac1{xy}$

Substitute
v=x^(-1) and
v'=-x^(-2)x' , and the resulting ODE is linear in
.

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

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

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

8.15

$yy'=-\frac x2$

This is separable; write
$y'=(\mathrm dy)/(\mathrm dx)$ , then

$y(\mathrm dy)/(\mathrm dx)=-\frac x2\implies y\,\mathrm dy=-\frac x2\,\mathrm dx$

Integrate both sides to get

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

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

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

so the particular solution is

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

10.15

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

$xy'''-y''+\frac1x=0$

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

$xv'-v=-\frac1x$

Divide through both sides by
x^2 :

$\frac{v'}x-\frac v{x^2}=\left(\frac vx\right)'=-\frac1{x^3}$

Integrate both sides and solve for
:

$\frac vx=\frac1{2x^2}+C$

$v=\frac1{2x}+Cx$

Solve for
by integrating both sides twice:

$y''=\frac1{2x}+Cx$

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

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

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 :

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

$-24a=2\implies a=-\frac1{12}$

$-24b=3\implies b=-\frac18$

Then the general solution is

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

Flowra answered Sep 25, 2021

by Flowra

6.6k points

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