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EDUARDO ESPINOZA. Análisis Matemático IV.

Variable separables.

1) [pic]

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4.- [pic]
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7.3extgydx + (1-ex)sec2ydy=0
Solución
[pic]dx + [pic]dy =0

[pic]dx + [pic]dy =[pic]
-3ln(1-ex) + ln|csc2y-ctg2y|=lnc

-ln(1-ex)3 + ln|csc2y-ctg2y|=lnc

-(1-ex)3 + |csc2y-ctg2y|=c

-(1-ex)3 + tgy= c

[pic]tgy = c + (1-ex)3 [pic]

PROBLEMA 12.

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PROBLEMA 13.
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14) [pic]
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17) [pic]

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[pic][pic][pic]

ejercicio 19
[pic]=[pic]
[pic]=[pic]
[pic] -[pic]
[pic]
[pic]arctg[pic]
arctg[pic]
[pic]
=arctg[pic]=ln[pic]

20. [pic]
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21) [pic]

[pic][pic]

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22.- (xy+x)dx=(x2y2+x2+y2+1)dy

Solución
(xy + x)dx-(x2y2 + x2 + y2 + 1)dy =0

x(y+1)dx – (y2+1)(x2+1)dy =0[pic]dx - [pic]dy =0

[pic]dx - [pic]dy =[pic]

[pic]dx - [pic]dy - [pic]dy=[pic]
[pic] - [pic] + 2(y+1) –ln|y+1| –ln|y+1| =2ln C
[pic] =[pic] - 2(y+1) +2ln|y+1| +2ln C
ln(x2+1) = y2 – 2y-3[pic]+ 4ln C + 4ln|y+1|
ln(x2+1) = y2 – 2y +4[pic]+ 4ln C + 4ln|y+1|
ln(x2+1) = y2 – 2y +4[pic] + 4ln|y+1|

ln(x2+1) = y2 – 2y +4[pic] + 4ln|y+1|

[pic]ln(x2+1) = y2 – 2y + 4[pic]

24)[pic]
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Por formula:
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Entonces:
a) [pic]

b) [pic]

c) [pic]

d) [pic]

e) [pic]

Respuesta:
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[pic][pic]
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Resolviendo:
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28. [pic]

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34.- y.Lnx. Lny. dx + dy=0
y.Lny. Lnx.dx+dy=0
[pic]=[pic]x.Lnx-x+Ln(Lny)=c

35)[pic]
Factorizar:
a)[pic]
Entonces:
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a)[pic]
b)[pic]
c)[pic]
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d)[pic]
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f) [pic]

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