# Separation of Variables Worksheet

Separation of Variables Worksheet
• Page 1
1.
Solve the equation $\frac{dy}{dx}$ = $\frac{3\left({y}^{2}-xy\right)}{xy-{x}^{2}}$.
 a. ${\left(\frac{y}{x}\right)}^{\frac{1}{2}}$ = C1 $x$ b. ${\left(\frac{x}{y}\right)}^{\frac{1}{2}}$ = C1 $x$ c. ${\left(\frac{x}{y}\right)}^{\frac{1}{2}}$ = C1 $x$ d. ${\left(\frac{y}{x}\right)}^{\frac{1}{2}}$ = C1 e. = C1 $x$

#### Solution:

dydx = 3(y2-xy)xy-x2
[Write the equation.]

dydx = 3y(y-x)x(y-x) = 3yx
[Factor both numerator and denominator.]

Let v = yx , y = vx

dydx = v + xdvdx

v + x dvdx = 3v

x dvdx = 2v

12v dv = 1x dx

1 / 21v dv = 1x dx
[Integrate on both sides.]

1 / 2ln |v| = ln |x| + C

ln |v12| = ln |x| + C

ln |v12| = ln |C1 x|
[C = ln |C1|.]

(yx)12 = C1 x
[Substitute v = yx.]

2.
Solve the equation $\frac{dy}{dx}$ = $\frac{x+y}{x-y}$.
 a. ln | $x$2 + $y$2 | = C b. 2 tan-1 ($\frac{y}{x}$) = C c. tan-1 ($\frac{y}{x}$) = ln |C1 ($x$2 + $y$2) | d. 2 tan-1($\frac{y}{x}$) = ln |C1 ($x$2 + $y$2) | e. tan-1 ($y$) = C1 ln | $x$2 + $y$2 |

#### Solution:

dydx = x+yx-y
[Write the differential equation.]

dydx = 1+yx1-yx
[Divide numerator and denominator by x.]

Put v = yx, y = vx

dydx = v + xdvdx

v + x dvdx = 1+v1-v
[Put v = yx.]

x dvdx = 1+v-v+v21-v

1-v1+v2 dv = 1x dx

1-v1+v2 dv = 1x dx
[Integrate on both sides.]

11+v2 dv - v1+v2 dv = 1x dx

tan-1 (v) - 1 / 2 ln | 1 + v2 | = ln | x | + C

2 tan-1 (v) = ln |x2| + ln | 1 + v2 | + C
[Multiply both sides by 2.]

2 tan-1 (v) = ln |C1 x2(1 + v2) |
[C = ln C1.]

2 tan-1 (yx) = ln |C1x2(1 + y2x2) |
[Replace v with yx.]

2 tan-1 (yx) = ln |C1 (x2 + y2) |

3.
Solve the equation $\mathrm{xy}$′ = $\mathrm{x + y}$.
 a. $y$ = $x$ ln | C1$x$ | b. $y$ = $\frac{1}{x}$ + C c. $x$ = $y$ ln | $x$ | + C d. $y$ = ln | C1$x$ | e. $y$ = $x$ + C

#### Solution:

y ′ = x+yx
[Write the differential equation.]

dydx = 1+yx1 = 1 + yx
[Divide numerator and denominator by x.]

Put v = yx, y = vx

dydx = v + xdvdx

v + x dvdx = 1 + v

x dvdx = 1

dv = 1x dx
[Integrate on both sides.]

v = ln | x | + C

v = ln | C1x |
[C = ln C1.]

yx = ln | C1x |
[Replace v with yx.]

y = x ln |C1x |

4.
Solve the equation $x$ $\mathrm{dy}$ - $y$ $\mathrm{dx}$ = $\sqrt{{x}^{2}+{y}^{2}}dx$.
 a. ($y$ + $\sqrt{{x}^{2}+{y}^{2}}$) = C1 $x$2 b. ($x$ + $\sqrt{{x}^{2}+{y}^{2}}$) = C1 $x$2 c. ($y$ + $\sqrt{{x}^{2}+{y}^{2}}$) = C1 $x$ d. $\mathrm{xy}$ = $c$1 e. ($\mathrm{xy}$ + $\sqrt{{x}^{2}+{y}^{2}}$) = C1 $x$

#### Solution:

x dy - y dx = x2+y2dx
[Write the differential equation.]

dydx = y+x2+y2x
[Find dydx.]

dydx = yx+1+y2x21
[Divide both numerator and denominator by x.]

Put v = yx, vx = y

v + x dvdx = dydx = v + 1+v2

x dvdx = 1+v2

dv1+v2 = 1x dx

dv1+v2 = 1x dx
[Integrate on both sides.]

ln | v + 1+v2 | = ln | x | + C

ln | v + 1+v2 | = ln | C1x |
[C = ln C1.]

ln | yx + 1+y2x2 | = ln | C1x |
[Replace v with yx.]

(y + x2+y2) = C1x2

5.
Solve $\frac{dy}{dx}$ = $\frac{xy}{{x}^{2}+{2y}^{2}}$.
 a. $y$2 = 4 ln | $y$ | + C b. $x$2 = 4 ln | $y$ | + C c. $y$2 = $y$2 ln | $y$ | + C d. $x$2 = 4$y$2 + C e. $\frac{{x}^{2}}{{4y}^{2}}$ = ln | $y$ | + C

#### Solution:

dydx = xyx2+2y2

dydx = yx1+2y2x2
[Divide both numerator and denominator by x2.]

Put v = yx, vx = y

v + x dvdx = dydx = v1+2v2

x dvdx = v1+2v2 - v

x dvdx = v-v-2v31+2v2

x dvdx = - 2v31+2v2

1+2v2-2v3 dv = 1x dx
[Integrate on both sides.]

- 1 / 2v-3 dv - 1v dv = 1x dx

14v2 - ln | v | = ln | x | + C

14v2 = ln | vx | + C

x24y2 = ln | y | + C
[Replace v with yx.]

6.
Solve $\frac{dy}{dx}$ = $\frac{{x}^{2}+xy+{y}^{2}}{{x}^{2}}$.
 a. tan-1 ($\frac{x}{y}$) = ln | $x$ | + C b. tan-1 ($\frac{y}{x}$) = $x$ + C c. tan-1 ($y$) = ln | $x$ | + C d. tan-1 ($\frac{y}{x}$) = ln | $x$ | + C e. tan-1 ($\frac{1}{x}$) = ln | $x$ | + C

#### Solution:

dydx = x2+xy+y2x2

dydx = 1+yx+y2x21
[Divide both numerator and denominator by x2.]

Put v = yx, vx = y

v + xdvdx = dydx

v + x dvdx = 1 + v + v2

x dvdx = 1 + v2

11+v2 dv = 1x dx
[Integrate on both sides.]

tan-1 (v) = ln | x | + C

tan-1 (yx) = ln | x | + C
[Replace v with yx.]

7.
Solve $\frac{dy}{dx}$ = $\frac{y}{x}+\frac{{y}^{2}}{{x}^{2}}$.
 a. $x$ + ln | $x$ | = C b. $\frac{y}{x}$ + ln | $x$ | = C c. $\frac{x}{y}$ + ln | $x$ | = C d. $\frac{x}{y}$ + $x$ = C e. $\frac{1}{y}$ + ln | $x$ | = C

#### Solution:

dydx = yx+y2x2
[Write the differential equation.]

Put v = yx, vx = y

v + xdvdx = dydx

v + x dvdx = v + v2

x dvdx = v2

1v2 dv = 1x dx

1v2 dv = 1x dx
[Integrate on both sides.]

- 1v + C = ln | x |

xy + ln | x | = C
[Replace v with yx.]

8.
Solve $y$ ′ = .
 a. ln | $x$ | - cos $y$ = C b. ln | $x$ | + sin ($\frac{y}{x}$) = C c. cos ($\frac{y}{x}$) - ln | $x$ | = C d. ln | $x$ | + cos ($\frac{x}{y}$) = C e. cos ($\frac{y}{x}$) + $x$ = C

#### Solution:

y ′ = yx - cosec yx
[Write the equation.]

dydx = yx - cosec yx

Put v = yx, vx = y

v + xdvdx = dydx

v + x dvdx = v - cosec v

x dvdx = - cosec v

- sin v dv = 1x dx
[Integrate on both sides.]

cos v = ln | x | + C

cos (yx) - ln | x | = C
[Replace v with yx.]

9.
Solve the homogeneous differential equation $x$ $\frac{dy}{dx}$ = $y$ + $x$ cos2 ($\frac{y}{x}$).
 a. $y$ = $x$ ln | C1$x$ | b. $y$ = $x$ tan-1 (ln | C1$x$ |) c. $x$ = $y$ tan-1 (ln |C1$x$ |) d. $y$ = $\frac{1}{x}$ tan-1 (ln | C1$x$ |) e. $y$ = tan-1 (ln | C1$x$ |)

#### Solution:

x dydx = y + x cos2 (yx)
[Write differential equation.]

dydx = yx + cos2 (yx)

Put v = yx, vx = y

v + xdvdx = dydx

v + x dvdx = v + cos2 (v)

1cos2v dv = 1x dx

sec2v dv = 1x dx
[Integrate on both sides.]

tan v = ln | x | + C

tan (yx) = ln | C1x |
[C = ln C1 and Replace v with yx.]

y = x tan-1 (ln | C1 x |)

10.
Solve the homogeneous differential equation $\frac{dy}{dx}$ = $\frac{y}{x}-{e}^{\frac{-y}{x}}$.
 a. ${e}^{\frac{x}{y}}$ = $\frac{1}{x}$ + C b. ${e}^{\frac{1}{x}}$ = ln | $\frac{{C}_{1}}{x}$ | c. ${e}^{y}$ = ln | $\frac{{C}_{1}}{x}$ | d. ${e}^{\frac{y}{x}}$ = ln | $\frac{{C}_{1}}{x}$ | e. ${e}^{\frac{x}{y}}$ = ln | $\frac{{C}_{1}}{x}$ |

#### Solution:

dydx = yx-e-yx
[Write the differential equation.]

Put v = yx, vx = y

v + xdvdx = dydx

v + x dvdx = v - e-v

x dvdx = - e-v

ev dv = - 1x dx

ev dv = - 1x dx
[Integrate both sides.]

ev = - ln | x | + C

ev = ln | 1x | + C

ev = ln | C1x |
[C = ln C1.]

eyx = ln | C1x |
[Replace v with yx.]