Please help with differential equations - QAmmunity.org

Please help with differential equations

asked Jul 27, 2021 155k views

1 Answer

4.15

$y'+\frac 2xy=x^3$

is a linear ODE; multiply both sides by the integrating factor
x^2 :

x^2y'+2xy=x^5

Now the left side can be condensed as the derivative of a product:

(x^2y)'=x^5

Integrate both sides and solve for
to get

$x^2y=\frac{x^6}6+C\implies y=\frac{x^4}6+\frac C{x^2}$

Given that
$y(1)=-\frac56$ , we find

$-\frac56=\frac16+C\implies C=-1$

and so the particular solution is

$\boxed{y(x)=\frac{x^4}6-\frac1{x^2}}$

5.15

$2y^2\,\mathrm dx+(x+e^(1/y))\,\mathrm dy=0$

You may be tempted to write this as an ODE in
y(x) , but the ODE in
x(y) is much easier to solve, since it's linear. Solve for
$(\mathrm dx)/(\mathrm dy)=x'$ and rearrange the terms:

$(\mathrm dx)/(\mathrm dy)=-(x+e^(1/y))/(2y^2)\implies x'+\frac x{2y^2}=-(e^(1/y))/(2y^2)$

Multiply both sides by the integrating factor
e^(-1/(2y)) , then solve for
:

e^(-1/(2y))x'+(e^(-1/(2y)))/(2y^2)x=-(e^(1/(2y)))/(2y^2)

$\left(e^(-1/(2y))x\right)'=-(e^(1/(2y)))/(2y^2)$

e^(-1/(2y))x=e^(1/(2y))+C

x=e^(1/y)+Ce^(1/(2y))

Given that
y=1 when
x=e , we have

$e=e+Ce^(1/2)\implies C=0$

so the particular solution is

$\boxed{x(y)=e^(1/y)}$

or, by solving for
,

$\boxed{y(x)=\frac1{\ln x}}$

6.15

y'-y=2xy^2

Dividing through both sides by
y^2 lets us write the equation in Bernoulli form:

y^(-2)y'-y^(-1)=2x

Substitute
v=y^(-1) , so that
v'=-y^(-2)y' . Then we get an ODE that is linear in
:

$-v'-v=2x\implies v'+v=-2x$

Multiply both sides by the integrating factor
e^x :

e^xv'+e^xv=(e^xv)'=-2xe^x

Integrate both sides and solve for
, then solve for
:

e^xv=-2e^x(x-1)+C

v=-2(x-1)+Ce^(-x)

$\frac1y=-2(x-1)+Ce^(-x)$

Given that
$y(0)=\frac12$ , we find

$2=2+C\implies C=0$

so the particular solution is

$\frac1y=-2(x-1)=2-2x\implies\boxed{y(x)=\frac1{2-2x}}$

Peter Sarnowski answered Aug 1, 2021

by Peter Sarnowski

4.9k points

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