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алмазия
@алмазия
July 2022
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Пожааааалуйста решите, с подробным решением, Умоляяяяю..
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Answers & Comments
simka105
Xy`+2y=1/x
y=z/x^2
y`=z`/x^2 - 2z/x^3
xy`+2y=z`/x - 2z/x^2+2z/x^2=z`/x
=1/x
z`/x
=1/x
z`
=1
z=x+с
y=z/x^2=(x+c)/x^2
*************************
y`*cos(x)-2y*sin(x)=2
y=z/cos^2(x)
y`=z`/cos^2(x)-2z/cos^3(x)*(-sin(x))
y`*cos(x)-2y*sin(x)=z`/cos(x)+2z/cos^2(x)*(sin(x))-2z/cos^2(x)*sin(x)=z`/cos(x)=2
z`/cos(x)=
2
z`=2*cos(x)
z=2sin(x)+C
y=z/cos^2(x)=(
2sin(x)+C)
/cos^2(x)
0 votes
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IUV
Verified answer
1) (x + C)/x^2
2)(c+2sin(x))*sec^2(x)
2 votes
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Answers & Comments
y=z/x^2
y`=z`/x^2 - 2z/x^3
xy`+2y=z`/x - 2z/x^2+2z/x^2=z`/x=1/x
z`/x=1/x
z`=1
z=x+с
y=z/x^2=(x+c)/x^2
*************************
y`*cos(x)-2y*sin(x)=2
y=z/cos^2(x)
y`=z`/cos^2(x)-2z/cos^3(x)*(-sin(x))
y`*cos(x)-2y*sin(x)=z`/cos(x)+2z/cos^2(x)*(sin(x))-2z/cos^2(x)*sin(x)=z`/cos(x)=2
z`/cos(x)=2
z`=2*cos(x)
z=2sin(x)+C
y=z/cos^2(x)=(2sin(x)+C)/cos^2(x)
Verified answer
1) (x + C)/x^22)(c+2sin(x))*sec^2(x)