EQUAÇÕES DIFERENCIAIS EM P

Universidade Católica de Pelotas

Centro Politécnico

Curso de Engenharia Elétrica

Calculo III

Exercício 1:

XP² – 3YP – X² = 0

3YP = XP² – X² /P

3Y = XP – X²/P

3P = P – 2X/P + (X +X²/P²)DP/DX => Y = DX/DY = P =>Of/Oy + Of/Op(DP/DY)

(-2P -2X/P) + (X + X²/P²)DP/DX = 0 *(DX)

(-2P -2X/P)DX + (X+X²/P²)DP = 0

-2P(1+X/P²)DX + X(1+X/P²)DP =0 => -2P e X em evidencia => /(1+X/P²)

-2PDX + XDP = 0 FI = 1/XP

-2DX/X + DP/P = 0

∫-2DX/X + ∫DP/P = 0

-2LnX + LnP = LnC

-LnX² + LnP = LnC

Ln P/X² = LnC /Ln

P/X² = C => P= X²C

{ P= X²C

{ XP² – 3YP – X² = 0

Substituindo em X:

X(X²C)² – 3Y(X²C) – X² = 0

C²X²X³ – 3CYX² – X² = 0 /X²

C²X³ – 3CY -1 =0

Exercício 2:

4YP² -2XP +Y =0 /(P)

4YP -2X +Y/P =0

2X = 4YP +Y/P

DX/DY = 1/P

2X = 2DX/DY = 2/P

APLICANDO DEHONDA:

2X = 4P + 1/P + (4Y – Y/P²)DP/DY

2/P – 1/P = 4P + (4Y – Y/P²)DP/DY *(P²)

P= 4P³ + (4P² – Y)DP/DY

(P -4P³)DY + Y(1-4P²)DP =0

P(1- 4P²)DY +Y(1-4P²)DP = 0 /(1-4P²)

PDY + YDP =0 FI = 1/PY

DY/Y + DP/P = 0

LnY + LnP = LnC

LnYP = LnC /(Ln)

YP=C

{ YP= C => P= C/Y

{ 4YP² -2XP +Y =0

Substituindo em Y:

4y(c/y)² – 2x(c/y) +y =0

4y(c²/y²) – 2xc/y + y = 0

4c²/y – 2cx/y + y = 0 4c = c = 2c

c²/y – cx/y + y =0 *(y)

c² – cx + y² = 0

Exercício 3:

XP² + YP – Y4 = 0 /P²

X + Y/P - Y4 /P² = 0

X = Y4 /P² - Y/P => X= DX/DY = 1/P

APLICANDO DEHONDA:

1/P = 4Y³/P² – 1/P +(-2Y4 /P³ + Y/P²)DP/DY => (Y4P-² – YP-¹) = [-2 * Y4P-²-¹ – (–1 * YP-¹-¹)]

-2/P + 4Y³/P² + (-2Y4 /P³ + Y/P²)DP/DY= 0 *P³

-2P² +4Y³P + (-2Y4 + YP)DP/DY = 0

EM EVIDENCIA:

-2P(P-2Y³) + Y(-2Y³+P)DP/DY = 0 * DY

-2P(P-2Y³)DY + Y(-2Y³+P)DP =0 /(-2Y³+P)

-2PDY + YDP = 0 FI = 1/PY

-2DY/Y + DP/P = 0

-2∫DY/Y + ∫DP/P = 0

-2LnY + LnP = LnC

-LnY² + LnP = LnC

Ln P/Y² = LnC /Ln

P/Y² = C P = CY²

{ P/Y²= C => P= CY²

{ XP² + YP – Y4 = 0

VOLTANDO EM Y:

X(CY²)² + Y(CY²) - Y4 = 0

C²XY4 + CY³- Y4 = 0 /Y³

C²XY + C- Y = 0 C²XY = C+ Y C = -C

SOLUÇÃO

Exercício 4:

Y²P² -2XYP +Y² -4 = 0 /YP

YP – 2X +Y/P - 4/YP = 0

2X = YP + Y/P – 4/YP => X= DX/DY = 1/P

2/P = YP + Y/P – 4/YP

APLICANDO DEHONDA:

2/P = P + 1/P + 4/Y²P +(Y -Y/P² +4/YP²)DP/DY => + YP-¹ = -1*YP-¹ -¹

2/P -1/P -P = 4/Y²P +(Y -Y/P² +4/YP²)DP/DY

1/P -P - 4/Y²P = (Y -Y/P² +4/YP²)DP/DY

(1/P -P - 4/Y²P)DY = (Y -Y/P² +4/YP²)DP *P²Y²

(PY² -P³Y² -4P)DY = (Y³P² – Y³ + 4Y)DP

(Y³P² – Y³ + 4Y)DP + (-PY² + P³Y² + 4P)DY = 0

EM EVIDENCIA:

Y(Y²P² – Y² + 4)DP +P(-Y² + P²Y² + 4)DY = 0 /(-Y² + P²Y² + 4)

YDP + PDY =0 FI = 1/YP

DP/P + DY/Y = 0

∫DP/P + ∫DY/Y = 0 LnP + LnY = LnC => LnPY = LnC /Ln

PY=C P = C/Y

{ PY= C => P= C/Y

{ Y²P² -2XYP +Y² -4 = 0

VOLTANDO EM Y:

Y²(C/Y)² -2XY(C/Y) +Y² – 4 = 0 Y²C²/Y² -2XCY/Y + Y² – 4 = 0 *(-1)

C² + 2CX -Y² + 4 =0 SOLUÇÃO -C² = C²