FIRST ORDER DIFFERENTIAL EQUATIONSassets.openstudy.com/updates/attachments/501902dfe... · FIRST ORDER DIFFERENTIAL EQUATIONS 3 3 (x2 y2)dx+(2xy)dy= 0 dy dx = y2 x2 2xy dy dx = y

FIRST ORDER DIFFERENTIAL EQUATIONS

FELIX HARVEY-ROSSER - 3457606

2C1

xdy

dx= y +

√x2 + y2

dy

dx=y

x+

√1 +

(yx

)2let

y

x= v

y = vx;dy

dx= v + x

dv

dx

v + xdv

dx= v +

√1 + v2

xdv

dx=√

1 + v2

dv√1 + v2

=dx

x∫dv√1 + v2

=

∫dx

x

let v = tanu; dv = sec2 udu;√

1 + v2 = secu∫secudu =

∫dx

x

ln (tanu+ secu) = lnx+ c

ln(v +

√1 + v2

)= lnx+ c

v +√

1 + v2 = ecx

y

x+

√1 +

(yx

)2= kx

y +√x2 + y2 = kx2

———1

2 FELIX HARVEY-ROSSER - 3457606

2

dy

dx=

4y2

x2+

5y

x+ 1

dy

dx= 4

(yx

)2+ 5

(yx

)+ 1

y

x= v; y = vx;

dy

dx= x

dv

dx+ v

xdv

dx+ v = 4v2 + 5v + 1

xdv

dx= 4v2 + 4v + 1

1

xdx =

dv

4v2 + 4v + 1

1

xdx =

dv

(2v + 1)2∫dx

x=

∫dv

(2v + 1)2

lnx+ c = −1

2· 1

2v + 1

(2v + 1) lnx+ ln ec = −1

2

−2(2v + 1) ln kx = 1

−2(2(yx

)+ 1)ln kx = 1

−2(2y + x) ln kx = x

x+ 2(2y + x) ln kx = 0

———

FIRST ORDER DIFFERENTIAL EQUATIONS 3

3

(x2 − y2)dx+ (2xy)dy = 0

dy

dx=y2 − x2

2xy

dy

dx=

( yx

)2 − 1

2( yx

)let

y

x= v

y = vx

dy

dx= v + x

dv

dx

v + xdv

dx=v2 − 1

2v

xdv

dx=v2 − 1

2v− v

xdv

dx=v2 − 1− 2v2

2v

xdv

dx= −1 + v2

2v

2v

1 + v2dv = −dx

x∫2v

1 + v2dv =

∫−dx

x

ln(1 + v2) = − lnx+ c

ln(1 + v2) = ln ec − lnx

ln(1 + v2) = lnk

x

1 + v2 =k

x

1 +(yx

)2=k

x

x2 + y2 = kx

———


x+y2

x= k

d

dx

(x+

y2

x

)=

d

dx(k)

1− y2

x2+ 2

y

x

dy

dx= 0

1 + 2y

x

dy

dx=y2

x2

x2 + 2xydy

dx= y2

2xydy

dx= y2 − x2

dy

dx=y2 − x2

2xy

———


4dy

dx+x+ y

x= 0

dy

dx= −x+ y

x

dy

dx= −

(1 +

y

x

)let v =

y

x

y = vx

dy

dx= v + x

dv

dx

v + xdv

dx= −(1 + v)

xdv

dx= −(1 + 2v)

dv

1 + 2v= −dx

x∫dv

1 + 2v=

∫−dx

x

1

2

∫2dv

1 + 2v= −

∫dx

x

1

2ln(1 + 2v) = − lnx+ c

1

2ln(1 + 2v) = ln ec − lnx

1

2ln(1 + 2v) = ln

k

x√1 + 2v =

k

x

1 + 2y

x=

(k

x

)2

x2 + 2xy = k2

———


x2 + 2xy = k2

d

dx

(x2 + 2xy

)=

d

dx

(k2)

2x+ 2y + 2xdy

dx= 0

2xdy

dx= −2(x+ y)

dy

dx= −(x+ y)

x

dy

dx+x+ y

x= 0

———


5

dy

dx=y − x+ 1

y + x+ 5

let y = Y + p; x = X + q

dy

dx=

dY

dX

dY

dX=

(Y + p)− (X + q) + 1

(Y + p) + (X + q) + 5

dY

dX=Y −X + (p− q + 1)

Y +X + (p+ q + 5)

p− q + 1 = 0 p+ q + 5 = 0

p = q − 1 (q − 1) + q + 5 = 0

2q + 4 = 0

q = −2p = (−2)− 1

p = −3Y = y − (−3) X = x− (−2)

dY

dX=Y −XY +X

dY

dX=

YX − 1YX + 1

letY

X= V

Y = V X

dY

dX= V +X

dV

dX

V +XdV

dX=V − 1

V + 1

XdV

dX=V − 1

V + 1− V

XdV

dX=V − 1

V + 1− V

(V + 1

V + 1

)X

dV

dX=V − 1− V (V + 1)

V + 1


XdV

dX= −V

2 + 1

V + 1

V + 1

V 2 + 1dV = −dX

X

∫V + 1

V 2 + 1dV = −

∫dX

X

1

2

∫2V

V 2 + 1dV +

∫1

V 2 + 1dV = − lnX + c

1

2ln (V 2 + 1) + tan−1 V = − lnX + c

1

2ln (V 2 + 1) + lnX + tan−1 V = c

1

2ln(X2(V 2 + 1

))+ tan−1 V = c

1

2ln

(X2

((Y

X

)2

+ 1

))+ tan−1

Y

X= c

1

2ln(Y 2 +X2

)+ tan−1

Y

X= c

1

2ln((y + 3)2 + (x+ 2)2

)+ tan−1

y + 3

x+ 2= c

1

2ln(y2 + 6y + 9 + x2 + 4x+ 4

)+ tan−1

y + 3

x+ 2= c

1

2ln(y2 + x2 + 6y + 4x+ 13

)+ tan−1

y + 3

x+ 2= c

———


6

dy

dx+

2x− y − 4

2y − x+ 5= 0

dy

dx=y − 2x+ 4

2y − x+ 5

let y = Y + p

x = X + q

dy

dx=

dY

dX

dY

dX=

(Y + p)− 2(X + q) + 4

2(Y + p)− (X + q) + 5

dY

dX=Y − 2X + (p− 2q + 4)

2Y −X + (2p− q + 5)

p− 2q + 4 = 0 2p− q + 5 = 0

p = 2q − 4 2(2q − 4)− q + 5 = 0

3q − 3 = 0

q = 1

p = 2(−1)− 4

p = −2

Y = y + 2 X = x− 1

dY

dX=Y − 2X

2Y −X

dY

dX=

YX − 2

2 YX − 1

letY

X= V ; Y = V X

dY

dX= V +X

dV

dX

V +XdV

dX=

V − 2

2V − 1

XdV

dX=

V − 2

2V − 1− V

XdV

dX=

V − 2

2V − 1− V

(2V − 1

2V − 1

)


XdV

dX=V − 2− V (2V − 1)

2V − 1

XdV

dX=V − 2− 2V 2 + V

2V − 1

XdV

dX= −2V

2 − V + 1

2V − 1

2V − 1

V 2 − V + 1dV = −2dX

X∫2V − 1

V 2 − V + 1dV = −2

∫dX

X

ln(V 2 − V + 1

)+ c = −2 lnX

ln((V 2 − V + 1)X2

)+ c = 0((

Y

X

)2

− Y

X+ 1

)X2 + ec = 0

Y 2 −XY +X2 + k = 0

(y + 2)2 − (x− 1)(y + 2) + (x− 1)2 + k = 0

y2 + 4y + 4− xy + y − 2x+ 2 + x2 − 2x+ 1 + k = 0

x2 + y2 − xy − 4x+ 5y + 7 + k = 0

———

x2 + y2 − xy − 4x+ 5y = −k′

d

dx

(x2 + y2 − xy − 4x+ 5y

)=

d

dx(k′)

2x+ 2ydy

dx− y − xdy

dx− 4 + 5

dy

dx= 0

2x− y − 4 = (x− 2y − 5)dy

dx

dy

dx=

2x− y − 4

x− 2y − 5

dy

dx+

2x− y − 4

2y − x+ 5= 0


7

dy

dx=x− y + 2

x− y + 1

let u = x− y + 1

y = x− u+ 1

dy

dx= 1− du

dx

1− du

dx=u+ 1

u

−du

dx=u+ 1

u− 1

−du

dx=u+ 1

u−(uu

)−du

dx=u+ 1− u

u

−du

dx=

1

u

−udu = dx

−∫udu =

∫dx

−u2

2= x+ c

−u2 − 2x = c2

−(x− y + 1)2 − 2x = c2

(y − x− 1)2 + 2x = c2

———


(y − x− 1)2 + 2x = c2

d

dx

((y − x− 1)2 + 2x

)=

d

dx(c2)

d

dx

(y2 − yx− y − xy + x2 + x− y + x+ 1 + 2x

)= 0

d

dx

(y2 − 2xy − 2y + x2 + 1 + 4x

)= 0

2ydy

dx− 2y − 2

dy

dx− 2x

dy

dx+ 2x+ 4 = 0

(2y − 2x− 2)dy

dx= 2y − 2x− 4

dy

dx=

2y − 2x− 2

2y − 2x− 4

dy

dx=x− y + 2

x− y + 1

———


8

dy

dx+

2x− y − 4

2x− y + 5= 0

let v = 2x− ydv

dx= 2− dy

dx

dy

dx= 2− dv

dx

2− dv

dx+v − 4

v + 5= 0

dv

dx= 2 +

v − 4

v + 5

dv

dx=

2(v + 5) + (v − 4)

v + 5

dv

dx=

3v + 6

v + 5

dx

dv=

v + 5

3(v + 2)

dx =v + 5

3(v + 2)dv

∫dx =

1

3

∫v + 5

v + 2dv

x =1

3

∫v

v + 2+

5

v + 2dv

x =1

3(v − 2 ln(v + 2) + 5 ln(v + 2)) + c1

x =1

3(v + 3 ln(v + 2)) + c1

x =(2x− y)

3+ ln((2x− y) + 2) + c1

x =(2x− y)

3+ ln(2x− y + 2) + c1


9

(x− 2y + 1)dx+ (4x− 3y − 6)dy = 0

dy

dx=

2y − x− 1

4x− 3y − 6

dY

dX=

2(Y + k)− (X + h)− 1

4(X + h)− 3(Y + k)− 6

dY

dX=

2Y −X + (2k − h− 1)

4X − 3Y + (4h− 3k − 6)

2k − h− 1 = 0 4h− 3k − 6 = 0

h = 2k − 1 4(2k − 1)− 3k − 6 = 0

5k − 10 = 0 k = 2 h = 2(2)− 1 h = 3

y = Y + 2 x = X + 3

Y = y − 2 X = x− 3

dY

dX=

2Y −X4X − 3Y

=2(YX

)− 1

4− 3(YX

)(Y

X

)= V Y = V X

dY

dX= X

dV

dX+ V

XdV

dX+ V =

2V − 1

4− 3V

XdV

dX=

2V − 1

4− 3V− V

XdV

dX=

2V − 1

4− 3V− V

(4− 3V

4− 3V

)X

dV

dX=

3V 2 − 2V − 1

4− 3V

dX

X=

4− 3V

3V 2 − 2V − 1dV∫

dX

X=

∫4− 3V

3V 2 − 2V − 1dV

lnX =

∫4− 3V

3V 2 − 2V − 1dV


3V 2 − 2V − 1 = (V − 1)

)3V 2 − 2V − 1

3V + 1

= (V − 1)

)3V 2 − 2V − 1

3V 2 − 3V

V − 1

V − 1

0

3V 2 − 2V − 1 = (V − 1)(3V + 1)

4− 3V

3V 2 − 2V − 1=

4− 3V

(V − 1)(3V + 1)=

α

(V − 1)+

β

(3V + 1)=

(3V + 1)α+ (V − 1)β

(V − 1)(3V + 1)

4− 3V = (3V + 1)α+ (V − 1)β

4− 3V = 3V α+ α+ V β − β

4− 3V = α− β + (3α+ β)V

4 = α− β

−3 = 3α+ β

4 + β = α

−3 = 3(4 + β) + β

−3 = 12 + 4β

−15 = 4β

β =−154

4 = α−(−154

)α = 4− 15

4=

16

4− 15

4=

1

4


lnX =

∫1/4

V − 1− 15/4

3V + 1dV

lnX =1

4

∫1

V − 1− 15

3V + 1dV

4 lnX = ln(V − 1)− 5 ln(3V + 1) + c

ln(X4)= ln

((V − 1)

(3V + 1)5

)+ c

X4 =V − 1

(3V + 1)5× ec

(3V + 1)5X4 = ec(V − 1)

(3V + 1)5X5 = AX(V − 1)

(3XV +X)5 = A(XV −X)

(3XY

X+X)5 = A(X

Y

X−X)

(3Y +X)5 = A(Y −X)

(3(y − 2) + (x− 3))5 = A((y − 2)− (x− 3))

(3y + x− 9)5 = A(y − x+ 1)

———


10

(5x+ 2y + 1)dx+ (2x+ y + 1)dy = 0

dy

dx= −5x+ 2y + 1

2x+ y + 1

dY

dX= −5(X + p) + 2(Y + q) + 1

2(X + p) + (Y + q) + 1

dY

dX= −5X + 2Y + (5p+ 2q + 1)

2X + Y + (2p+ q + 1)

2p+ q + 1 = 0 5p+ 2q + 1 = 0

q = −2p− 1 5p+ 2(−2p− 1) + 1 = 0

q = −2(1)− 1 p− 1 = 0

q = −3 p = 1

y = Y − 3 x = X + 1

Y = y + 3 X = x− 1

dY

dX= −5X + 2Y

2X + Y

dY

dX= −

5 + 2 YX

2 + YX

Y

X= V Y = V X

dY

dX= X

dV

dX+ V

XdV

dX+ V = −5 + 2V

2 + V

XdV

dX= −V 2 + V

2 + V− 5 + 2V

2 + V

XdV

dX= −V (2 + V ) + 5 + 2V

2 + V

XdV

dX= −V

2 + 4V + 5

2 + V


1

XdX = − 2 + V

V 2 + 4V + 5dV∫

1

XdX = −

∫V + 2

V 2 + 4V + 5dV

lnX = −∫

V + 2

V 2 + 4V + 5dV

lnX = −1

2

∫2V + 4

V 2 + 4V + 5dV

lnX = −1

2ln(V 2 + 4V + 5) + c

X = ec(V 2 + 4V + 5)−1/2

X(V 2 + 4V + 5)1/2 = A

X2(V 2 + 4V + 5) = A2

X2

(Y

X

)2

+ 4X2

(Y

X

)+ 5X2 = B

Y 2 + 4XY + 5X2 = B

(y + 3)2 + 4(x− 1)(y + 3) + 5(x− 1)2 = B

(y2 + 6y + 9) + 4(xy + 3x− y − 3) + 5(x2 − 2x+ 1) = B

y2 + 6y + 9 + 4xy + 12x− 4y − 12 + 5x2 − 10x+ 5 = B

y2 + 2y + 4xy + 5x2 + 2x+ 2 = B

———

d

dx

(y2 + 2y + 4xy + 5x2 + 2x+ 2

)=

d

dx(B)

(2y + 2 + 4x)dy

dx+ 4y + 10x+ 2 + 0 = 0

(2y + 2 + 4x)dy

dx= −(10x+ 4y + 2)

(y + 1 + 2x)dy = −(5x+ 2y + 1)dx

(5x+ 2y + 1)dx+ (2x+ y + 1)dy = 0


11

(3x− y + 1)dx− (6x− 2y − 3)dy = 0

(3x− y + 1)dx = (6x− 2y − 3)dy

dy

dx=

3x− y + 1

6x− 2y − 3

let w = 3x− y; dw

dx= 3− dy

dx

dy

dx= 3− dw

dx

3− dw

dx=

w + 1

2w − 3

dw

dx= 3− w + 1

2w − 3

dw

dx=

3(2w − 3)− (w + 1)

2w − 3

dw

dx=

6w − 9− w − 1

2w − 3

dw

dx=

5w − 10

2w − 3

dw

dx= 5

w − 2

2w − 3

5dx =2w − 3

w − 2dw∫

5dx =

∫2w − 3

w − 2dw

5

∫dx =

∫2(w − 2) + 1

w − 2dw

5x =

∫2 +

1

w − 2dw

5x = 2w + ln(w − 2) + c

5x = 2(3x− y) + ln(3x− y − 2) + c

5x = 6x− 2y + ln(3x− y − 2) + c

0 = x− 2y + ln(3x− y − 2) + c


———

2y = x+ ln(3x− y − 2) + c

d

dx(2y) =

d

dx(x+ ln(3x− y − 2) + c)

2dy

dx= 1 +

3− dydx

3x− y − 2

2dy

dx=

3x− y + 1

3x− y − 2−

dydx

3x− y − 2

2dy

dx+

dydx

3x− y − 2=

3x− y + 1

3x− y − 2

dy

dx

(2 +

1

3x− y − 2

)=

3x− y + 1

3x− y − 2

dy

dx

(2(3x− y − 2) + 1

3x− y − 2

)=

3x− y + 1

3x− y − 2

dy

dx

(6x− 2y − 3

3x− y − 2

)=

3x− y + 1

3x− y − 2

dy

dx(6x− 2y − 3) = (3x− y + 1)

(3x− y + 1)dx = (6x− 2y − 3)dy

———


12

dy

dx= −2x+ 3y + 1

4x+ 6y + 1

let v = 2x+ 3y;dv

dx= 2 + 3

dy

dx

dy

dx=

1

3

(dv

dx− 2

)1

3

(dv

dx− 2

)= − v + 1

2v + 1

dv

dx= 2− 3(v + 1)

2v + 1

dv

dx=

2(2v + 1)− 3(v + 1)

2v + 1

dv

dx=

4v + 2− 3v − 3

2v + 1

dv

dx=

v − 1

2v + 1

dx =2v + 1

v − 1dv∫

dx =

∫2v + 1

v − 1dv

let w = v − 1; w + 1 = v; dw = dv∫dx =

∫2(w + 1) + 1

wdw∫

dx =

∫2 +

3

wdw

x = 2w + 3 ln(w) + c

x = 2(v − 1) + 3 ln(v − 1) + c

x = 2v − 2 + 3 ln(v − 1) + c

x = 2(2x+ 3y − 1) + 3 ln(2x+ 3y − 1) + c

x = 4x+ 6y + 3 ln(2x+ 3y − 1)− 2 + c

0 = 3x+ 6y + 3 ln(2x+ 3− 1)− 2 + c


———

0 = 3x+ 6y + 3 ln(2x+ 3− 1)− 2 + c

d

dx(0) =

d

dx(3x+ 6y + 3 ln(2x+ 3y − 1)− 2 + c)

0 = 3 + 6dy

dx+ 3

2 + 3dydx

2x+ 3y − 1

0 = 3

(1 + 2

dy

dx+

2 + 3dydx

2x+ 3y − 1

)

0 =

(1 +

2

2x+ 3y − 1

)+

(2 +

3

2x+ 3y − 1

)dy

dx

(2 +

3

2x+ 3y − 1

)dy

dx= −

(1 +

2

2x+ 3y − 1

)(2(2x+ 3y − 1) + 3

2x+ 3y − 1

)dy

dx= −

(2x+ 3y − 1 + 2

2x+ 3y − 1

)(4x+ 6y + 1

2x+ 3y − 1

)dy

dx= −

(2x+ 3y + 1

2x+ 3y − 1

)

dy

dx= −

(2x+ 3y + 1

2x+ 3y − 1

)(2x+ 3y − 1

4x+ 6y + 1

)

dy

dx= −

(2x+ 3y + 1

4x+ 6y + 1

)

———

FIRST ORDER DIFFERENTIAL EQUATIONSassets.openstudy.com/updates/attachments/501902dfe... · FIRST ORDER DIFFERENTIAL EQUATIONS 3 3 (x2 y2)dx+(2xy)dy= 0 dy dx = y2 x2 2xy dy dx = y

Text of FIRST ORDER DIFFERENTIAL EQUATIONSassets.openstudy.com/updates/attachments/501902dfe... · FIRST...

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11.differential equations final - Sakshieducation.com ...sakshieducation.com/EAMCET/mathematics/Bitbanks/Calculus/... · 2 2 3 / 2 2 / 1 d y dx dx dy ... 1 −x 2 dx −1 −y2 dy

· Exercise 5.3 2x + 3y = sin x Differentiating both sides with respect to x, we get — sin x dx dx dx dy cosx — 2 dx 2x + = sin y dy dx cos x Question 2: 2x + = sin y Answer

COMPUTER BASED TEST (CBT) Memory Based Questions & … · x2= 4b(y + b), tgk¡ b ,d izkpy gS] dk vody lehdj.k gksxk % (1) 2 2 x dx dy 2 dx dy x (2) x dx dy 2y dx dy x 2 (3) 2 2 x

Ordinary Differential Equations - Numerical FactoryOrdinary Differential Equations • Analytical Solution: (only in few cases) • Separate variables: dy dx x dy x dx y x C ³ ³

Ejercicios)sgpwe.izt.uam.mx/.../Ejercicios_Segundo_Parcial.pdf · 7. Suponga que y (a) Si dx/dt = 3, encuentre dy/dt cuando x = 4. (b) Si dy/dt = 5, encuentre dx/dt cuando x = 12

Analisi Matematica 2 - Altervistammarras.altervista.org/Mat2_Lez5_int_Doppi.pdf2 1 x 3e y x dx dy oppure Z 2 1 Z 1 0 x 3e y x dy dx ? Essendo R 1 0 x 3e y x dy un integrale immediato

· 6. dy dx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ T = T(2−3T) (1−T)3 When T = 1 2 dy dx ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 5= 1 2 2−3 ( 2) 1−1 ( 2) 1− 3 ( 2) = 1

Answers - Nulake · 36 99 36 13 Page 38 186. dy dx = 2x cos x – x2 sin x 187. dy dx = 3 tan 3x + (9x – 6) sec2 3x 188. g’(x) = dy dx = 2x cot(5x – 1)

2 Movimento Retilíneo - sistemas.eel.usp.brsistemas.eel.usp.br/docentes/arquivos/2166002/LOB1018/02-Movimento... · d y z dy dz dx dx dx d yz dz dy yz dx dx dx dy dy dx dt dx dt

eduwavepool.unizwa.edu.om€¦ · Use dy/dx = I/(dx/dy). (y — 2r)ez = 2 — y + 21, dx/dÿ = —2 + ÿ — 21 is linear, dx/dÿ + 2x = — 2, x = + y/2 — 5/4 u = c, uxdx + uydy

Computational Graph & Backpropagationspeech.ee.ntu.edu.tw/~tlkagk/courses/MLDS_2018/Lecture/Graph.pdfDNN%20backprop.ecm.mp4/index.html •Simple version: ... yx dx dy dy dz dx dz 'xyz

Teil I GANZRATIONALE FUNKTIONEN · 2014. 2. 24. · Differenzialquotient - Ableitungen - 2 - ABLEITUNGSREGELN sind Regeln zur Bestimmung des Differenzialquotienten dy dx. dy y dx

MPPT of Photovoltaic Systems using Extremum–Seeking · =Ksign µ dy dx : (6) Note that the algorithm measures the sign of dy=dt, whereas the resulting dynamics are governed by dy=dx

w u dy dx dy wy - University of Texas at · PDF fileB (6a) x (6b) y y dy xx V xy V yx V V yy dx x dx x xy xy w w V V dx x xx xx w w V V dy y yy yy w V V dy y yx yx w w V V tor per

Drill: Find dy / dx

156 Section 4.1 Chapter 4 () More Derivatives solns.pdf · dy dy du dx du dx = 62 6 6 662 (cos ) ( ) (sin)() sin sin ( ) dd ux du dx u u x =+ =− =− =− + (b) dy dy du dx du dx

dx x - University of Colorado Boulder · PDF file · 2014-12-22... nd dy=dx. (a) y= 3x3 + 2x2 p x Answer: dy dx = 15 2 ... Find the value of the integral: R 7 0 f0(x)dx Answer: R

An integral lift of contact homology - Columbia Universitymath.columbia.edu/~nelson/2017_penn.pdf · d = dy ^dx = dx ^dy) ^d = dx ^dy ^dz Jo Nelson An integral lift of contact homology

THE LOCAL EQUIVALENCE PROBLEM FOR d2y/dx = F(x, y, dy/dx

Xe .ie :±÷÷÷¥÷÷÷÷¥÷ii¥÷...dy y xy dx x xy y = 7) sin sin tan tan cos cos dy x y xy dx x y == 8) 1cos sin y y dy y e x dx xe x + = 9) 2 2 x y x y dy y y e dx yxe = 10)

CMSC 330: Organization of Programming Languages · letdy= ey-. syin sqrt(dx . dx +. dy. dy)OCaml Rust Can include patterns in parameters directly, too We’ll see Rust structslater

Uplift Education / Overview · Web view2012. 10. 3. · 2. derivative: d 2 y d 2 x = d dx dy dx = d dθ dy dx dθ dx = d dθ dy dx dx dθ = 1 dx dθ d dθ dr dθ sin θ + rcos θ

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FIRST ORDER DIFFERENTIAL EQUATIONSassets.openstudy.com/updates/attachments/501902dfe... · FIRST ORDER DIFFERENTIAL EQUATIONS 3 3 (x2 y2)dx+(2xy)dy= 0 dy dx = y2 x2 2xy dy dx = y

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