Question

Variable Separable: Give the particular and general solution

1. dr/dt = —4rt;

when t =0, r = 0.

2. 2xyy’ =1+y?

when x = 2, y =3

3. xyy’=1+y?

when x =2, y =3.

4. 2ydx = 3xdy

When x =2, y= 1.

5. 2ydx = 3xdy

When x = —2, y= 1.

6. 2ydx = 3xdy;

When x =2, y= —l.

7. y' =x^ (y— x^2);

when x= 0, y =0.

8. xy^2 dx +e^x dy =0;

when x=∞, y=1/2

9. (2a^2 —r^2) dr = r3 sin Ɵ dƟ

When Ɵ =0, r =a.

10.v(dv/dx) = g

when x = xo, v = vo

Answer 1

Solution.

1)

$\frac{dr}{dt}=-4rt \newline \frac{dr}{r}=-4tdt\newline \ln|r|=-2t^2+C \newline r=C_1e^{-2t^2}$ general solution

When $t=0, r=0,$ then $C_1=0.$

$r=0$ particular solution.

2)

$2xyy'=1+y \newline \frac{y}{y+1}dy=\frac{dx}{2x} \newline y+1-\ln{|y+1|}=\frac{1}{2}\ln{|x|}+C \newline y-\ln{|y+1|}=\frac{1}{2}\ln{|x|}+C_1 \newline$ general solution

When $x=2, y=3,$ then

$3-\ln{4}=\ln{\sqrt{2}}+C_1, \newline C_1=3-\ln{4}-\ln{\sqrt{2}}=3-\ln{\sqrt{32}}.$

$y-\ln{|y+1|}=\frac{1}{2}\ln{|x|}+3-\ln{\sqrt{32}}$ particular solution.

3)

$xyy'=1+y \newline \frac{y}{y+1}dy=\frac{dx}{x} \newline y+1-\ln{|y+1|}=\ln{|x|}+C \newline y-\ln{|y+1|}=\ln{|x|}+C_1 \newline$ general solution

When $x=2, y=3,$ then

$3-\ln{4}=\ln{2}+C_1, \newline C_1=3-\ln{4}-\ln{2}=3-\ln{8}.$

$y-\ln{|y+1|}=\ln{|x|}+3-\ln{8}$ particular solution.

4)

$2ydx=3xdy \newline 2\frac{dx}{x}=3\frac{dy}{y} \newline 2\ln{|x|}=3\ln{|y|}+C \newline y=C_1x^{\frac{2}{3}}$ general solution

When $x=2, y=1,$ then

$1=C_12^{\frac{2}{3}}, C_1=\frac{1}{\sqrt[3]{4}}=\frac{\sqrt[3]{2}}{2}.$

$y=\frac{\sqrt[3]2}{2}x^{\frac{2}{3}}$ particular solution.

5)

$2ydx=3xdy \newline 2\frac{dx}{x}=3\frac{dy}{y} \newline 2\ln{|x|}=3\ln{|y|}+C \newline y=C_1x^{\frac{2}{3}}$ general solution

When $x=-2, y=1,$ then

$1=C_1(-2)^{\frac{2}{3}}, C_1=-\frac{1}{\sqrt[3]{4}}=-\frac{\sqrt[3]{2}}{2}.$

$y=-\frac{\sqrt[3]2}{2}x^{\frac{2}{3}}$ particular solution.

6)

$2ydx=3xdy \newline 2\frac{dx}{x}=3\frac{dy}{y} \newline 2\ln{|x|}=3\ln{|y|}+C \newline y=C_1x^{\frac{2}{3}}$ general solution

When $x=2, y=-1,$ then

$-1=C_12^{\frac{2}{3}}, C_1=-\frac{1}{\sqrt[3]{4}}=-\frac{\sqrt[3]{2}}{2}.$

$y=-\frac{\sqrt[3]2}{2}x^{\frac{2}{3}}$ particular solution.

7)

$y'=x^{y-x^2}$

This is not an equation with separate variables.

Maybe there is an error in the task condition?

8)

$xy^2dx+e^xdy=0 \newline \frac{xdx}{e^x}=-\frac{dy}{y^2} \newline -xe^{-x}-e^{-x}=\frac{1}{y}+C \newline y=-\frac{e^x}{x+1+Ce^x}$ general solution

When $x=\infty, y=\frac{1}{2},$ then

$C=-{2}.$

$y=-\frac{e^x}{x+1-2e^x}$ particular solution.

9)

$(2a^2-r^2)dr=r^3\sin{\theta}d\theta \newline \frac{2a^2-r^2}{r^3}dr=\sin{\theta}d\theta \newline 2a^2\int r^{-3}dr-\int \frac{dr}{r}=\int \sin{\theta}d\theta \newline -\frac{a^2}{r^2}-\ln{|r|}=-\cos{\theta}+C \newline \ln{|r|}=\cos{\theta}-\frac{a^2}{r^2}+C_1 \newline$ general solution

When $\theta=0, r=a,$ then

$C_1=\ln{|a|}-1+1=\ln{|a|}.$

$\ln{|r|}=\cos{\theta}-\frac{a^2}{r^2}+\ln{|a|}.$

10)

$v\frac{dv}{dx}=g \newline vdv=gdx \newline \frac{v^2}{2}=gx+C \newline v^2=2gx+C_1$ general solution

When $x=x_0, v=v_0,$ then

$C_1=v_0^2-2gx_0.$

$v^2=2gx+v_0^2-2gx_0, \text{ or} \newline v=\pm\sqrt{2gx+v_0^2-2gx_0}$ particular solution.