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일반적인 미적분학 문제
D(t)=(100ln(24+2))/(24+2)
D
(
t
)
=
1
0
0
ln
(
2
4
+
2
)
2
4
+
2
y^{''}+y^'-y=sin(x)
y
′
′
+
y
′
−
y
=
sin
(
x
)
y^{''}+y^'=4t
y
′
′
+
y
′
=
4
t
y^{''}-y+t^2=0
y
′
′
−
y
+
t
2
=
0
y^{'''}+y^{''}=4x^2+3e^x
y
′
′
′
+
y
′
′
=
4
x
2
+
3
e
x
y^{''''}+2y^{''}+y=3x+4
y
′
′
′
′
+
2
y
′
′
+
y
=
3
x
+
4
y^{''}-18y^'+81y=480sin(x)-210cos(x)
y
′
′
−
1
8
y
′
+
8
1
y
=
4
8
0
sin
(
x
)
−
2
1
0
cos
(
x
)
y^{''}+9y=2,y^'(0)=5
y
′
′
+
9
y
=
2
,
y
′
(
0
)
=
5
y^{''}+y^'-12y=sin(t)
y
′
′
+
y
′
−
1
2
y
=
sin
(
t
)
y^{''}-y^'-6y=250te^{-2t}
y
′
′
−
y
′
−
6
y
=
2
5
0
te
−
2
t
y^{''}-8y^'+20y=100x^2
y
′
′
−
8
y
′
+
2
0
y
=
1
0
0
x
2
y^{''}-3y^'+2y=x^2+x
y
′
′
−
3
y
′
+
2
y
=
x
2
+
x
y^{''}-8y^'+16y=(e^{4x})/x
y
′
′
−
8
y
′
+
1
6
y
=
e
4
x
x
(2D^2-3D+1)y=csc(2pix)
(
2
D
2
−
3
D
+
1
)
y
=
csc
(
2
π
x
)
solvefor d,d(x)=(x^2+13x+42)/(x^2-3x-28)
solvefor
d
,
d
(
x
)
=
x
2
+
1
3
x
+
4
2
x
2
−
3
x
−
2
8
y^{''}-7y^'=x^2
y
′
′
−
7
y
′
=
x
2
x^{''}+3x^'+2x=3cos(2t)
x
′
′
+
3
x
′
+
2
x
=
3
cos
(
2
t
)
2x^{''}+5x^'+4x=10r(t)
2
x
′
′
+
5
x
′
+
4
x
=
1
0
r
(
t
)
y^{'''}-3y^{''}+3y^'-y=e^x-x^3+16
y
′
′
′
−
3
y
′
′
+
3
y
′
−
y
=
e
x
−
x
3
+
1
6
x^{''}+6x^'-4x=8e^t
x
′
′
+
6
x
′
−
4
x
=
8
e
t
x^{''}+4x^'+43x=9cos(6t)
x
′
′
+
4
x
′
+
4
3
x
=
9
cos
(
6
t
)
y^{''}+9y=2cos(3x)-3sin(3x)
y
′
′
+
9
y
=
2
cos
(
3
x
)
−
3
sin
(
3
x
)
y^{''}-4y^'+3=0
y
′
′
−
4
y
′
+
3
=
0
y^{''}+3y=6x
y
′
′
+
3
y
=
6
x
y^{''}-3y^'+2=0
y
′
′
−
3
y
′
+
2
=
0
-2y^{''}+7y^'-y-8=-6x^2-2x^3e^x
−
2
y
′
′
+
7
y
′
−
y
−
8
=
−
6
x
2
−
2
x
3
e
x
y^{''}-y^'-2y=4t^2
y
′
′
−
y
′
−
2
y
=
4
t
2
(d^2y)/(dx^2)+25y=5x^2+x
d
2
y
dx
2
+
2
5
y
=
5
x
2
+
x
25y^{''}-9y=sin(x)
2
5
y
′
′
−
9
y
=
sin
(
x
)
y^{''}+2y^'-24y=6x
y
′
′
+
2
y
′
−
2
4
y
=
6
x
y^{''}+3y^'+5y=sin(t)
y
′
′
+
3
y
′
+
5
y
=
sin
(
t
)
y^{''}-3y^'+2y=xcos(x)
y
′
′
−
3
y
′
+
2
y
=
x
cos
(
x
)
y^{''}+y=sec(x)csc(x)
y
′
′
+
y
=
sec
(
x
)
csc
(
x
)
y^{'''}+3y^{''}-4y=e^{-2x}
y
′
′
′
+
3
y
′
′
−
4
y
=
e
−
2
x
y^{''}+y^'-2y+1=0
y
′
′
+
y
′
−
2
y
+
1
=
0
(d^2y)/(dx^2)-4(dy)/(dx)-12=3cos(x)
d
2
y
dx
2
−
4
dy
dx
−
1
2
=
3
cos
(
x
)
y^{''}+4y=-1
y
′
′
+
4
y
=
−
1
y^{''}-3y^'+2y=(e^{2x})/(1+e^{2x)}
y
′
′
−
3
y
′
+
2
y
=
e
2
x
1
+
e
2
x
(d^2-2d+2)y=4x-2+2e^xsin(x)
(
d
2
−
2
d
+
2
)
y
=
4
x
−
2
+
2
e
x
sin
(
x
)
y^{''}-12y^'+36y=36x+4
y
′
′
−
1
2
y
′
+
3
6
y
=
3
6
x
+
4
y^{''}-2y^'=30e^{-3t},y(0)=1,y^'(0)=0
y
′
′
−
2
y
′
=
3
0
e
−
3
t
,
y
(
0
)
=
1
,
y
′
(
0
)
=
0
2y^{''}+ay^'+ay=e^{-2x}+x
2
y
′
′
+
ay
′
+
ay
=
e
−
2
x
+
x
y^{''}+2y^'+2y=4e^{-x}sec^3(x)
y
′
′
+
2
y
′
+
2
y
=
4
e
−
x
sec
3
(
x
)
y^{''}+36y=36sec^2(6t)
y
′
′
+
3
6
y
=
3
6
sec
2
(
6
t
)
y^{''}-3y^'=-4-6x
y
′
′
−
3
y
′
=
−
4
−
6
x
y^{''}+y^'-2y=1+t,y(0)=2,y^'(0)=0
y
′
′
+
y
′
−
2
y
=
1
+
t
,
y
(
0
)
=
2
,
y
′
(
0
)
=
0
y^{''}+y=2tan(x)
y
′
′
+
y
=
2
tan
(
x
)
1/2 x^{''}+1/2 x^'-6x=10cos(3t)
1
2
x
′
′
+
1
2
x
′
−
6
x
=
1
0
cos
(
3
t
)
y^{''}+4y^'+4y=4x+5
y
′
′
+
4
y
′
+
4
y
=
4
x
+
5
y^{''}+4y^'+4y=4x+6
y
′
′
+
4
y
′
+
4
y
=
4
x
+
6
y^{''}-2y^'-8y=-16t+16t^2
y
′
′
−
2
y
′
−
8
y
=
−
1
6
t
+
1
6
t
2
y^{''}-16y=48t-24e^{-4t}
y
′
′
−
1
6
y
=
4
8
t
−
2
4
e
−
4
t
x^{''}+2x^'+50x=2cos(7t)
x
′
′
+
2
x
′
+
5
0
x
=
2
cos
(
7
t
)
y^{''}+y=(cos(x))^2
y
′
′
+
y
=
(
cos
(
x
)
)
2
(d^2y)/(dt^2)+4y=4sec^2(2t)
d
2
y
dt
2
+
4
y
=
4
sec
2
(
2
t
)
y^{''}-2y^'+y=e^{-t}ln(t)
y
′
′
−
2
y
′
+
y
=
e
−
t
ln
(
t
)
y^{''}+2y^'+10y=tcos(t)+e^t
y
′
′
+
2
y
′
+
1
0
y
=
t
cos
(
t
)
+
e
t
y^{''}+5y^'+6y=-19te^{2t}
y
′
′
+
5
y
′
+
6
y
=
−
1
9
te
2
t
y^{''}+1/2 y^'+y=cos(t)
y
′
′
+
1
2
y
′
+
y
=
cos
(
t
)
y^{'''}-4y^'=3x-1
y
′
′
′
−
4
y
′
=
3
x
−
1
y^{''}-y^'+361y=19sin(19t)
y
′
′
−
y
′
+
3
6
1
y
=
1
9
sin
(
1
9
t
)
y^{''}-y^'=2e^x
y
′
′
−
y
′
=
2
e
x
5y^{''}+25y^'+1/(10^{-2)}y=10e^{-0/4}
5
y
′
′
+
2
5
y
′
+
1
1
0
−
2
y
=
1
0
e
−
0
4
y^{''}+2y^'=2x+4
y
′
′
+
2
y
′
=
2
x
+
4
y^{''}+4y^'+3y=e^t+sin(t)
y
′
′
+
4
y
′
+
3
y
=
e
t
+
sin
(
t
)
y^{''}-2y^'-3y=2e^{3t},y(0)=1,y^'(0)=0
y
′
′
−
2
y
′
−
3
y
=
2
e
3
t
,
y
(
0
)
=
1
,
y
′
(
0
)
=
0
y^{''''}-y=5t+cos(t)
y
′
′
′
′
−
y
=
5
t
+
cos
(
t
)
y^{''}+y=t,y(0)=1,y^'(0)=2
y
′
′
+
y
=
t
,
y
(
0
)
=
1
,
y
′
(
0
)
=
2
y^{''}-6y^'+34=0
y
′
′
−
6
y
′
+
3
4
=
0
y^{''}+25y=5sec(5t)
y
′
′
+
2
5
y
=
5
sec
(
5
t
)
y^{''}+cy-c-k=0
y
′
′
+
cy
−
c
−
k
=
0
y^{''}+4y^'+4y=e^{3x}
y
′
′
+
4
y
′
+
4
y
=
e
3
x
y^{''}+y=3sin(2x)+xcos(2x)
y
′
′
+
y
=
3
sin
(
2
x
)
+
x
cos
(
2
x
)
(d^2y)/(dx^2)-8(dy)/(dx)+159=0
d
2
y
dx
2
−
8
dy
dx
+
1
5
9
=
0
y^{''}+3y^'+3y=6sin(2t)
y
′
′
+
3
y
′
+
3
y
=
6
sin
(
2
t
)
y^{''}-14y^'+49y=14x+3
y
′
′
−
1
4
y
′
+
4
9
y
=
1
4
x
+
3
y^{'''}+y^{''}-y^'-1=2(1+e^{-x})
y
′
′
′
+
y
′
′
−
y
′
−
1
=
2
(
1
+
e
−
x
)
y^{''}+2y^'=e^{-2x}
y
′
′
+
2
y
′
=
e
−
2
x
y^{''}+4y= 2/(sin(2x))
y
′
′
+
4
y
=
2
sin
(
2
x
)
x^{''}+3x^'+2x=5cos(t)+sin(2t)
x
′
′
+
3
x
′
+
2
x
=
5
cos
(
t
)
+
sin
(
2
t
)
3x^{''}+53x=5cos(4t)
3
x
′
′
+
5
3
x
=
5
cos
(
4
t
)
y^{''}-y^'-2y=3cos(3x)
y
′
′
−
y
′
−
2
y
=
3
cos
(
3
x
)
y^{''}-6y^'+8y=e^{ax}
y
′
′
−
6
y
′
+
8
y
=
e
ax
25y^{''}-10y^'-2y=4e^t
2
5
y
′
′
−
1
0
y
′
−
2
y
=
4
e
t
y^{''}+0.5y^'+y=cos(t)
y
′
′
+
0
.
5
y
′
+
y
=
cos
(
t
)
y^{''}+y=2cos(x),y(0)=1,y^'(0)=0
y
′
′
+
y
=
2
cos
(
x
)
,
y
(
0
)
=
1
,
y
′
(
0
)
=
0
y^{''}-3y^'+y=2sin(2x)
y
′
′
−
3
y
′
+
y
=
2
sin
(
2
x
)
(D+1)^3y=16(2x+3)^{-1}e^x
(
D
+
1
)
3
y
=
1
6
(
2
x
+
3
)
−
1
e
x
y^{''}-4y^'-8y=8x^2+8x+1
y
′
′
−
4
y
′
−
8
y
=
8
x
2
+
8
x
+
1
y^{''}-9y=63
y
′
′
−
9
y
=
6
3
solvefor D,D(x)=e^{2x^3+4}
solvefor
D
,
D
(
x
)
=
e
2
x
3
+
4
y^{''}+3y^'-4y=12
y
′
′
+
3
y
′
−
4
y
=
1
2
y^{''}+2y^'+y=x^2e^x
y
′
′
+
2
y
′
+
y
=
x
2
e
x
y^{''}-6y^'+5y=-1+5x
y
′
′
−
6
y
′
+
5
y
=
−
1
+
5
x
y^{''}-6y^'+9y=8e^{3x}
y
′
′
−
6
y
′
+
9
y
=
8
e
3
x
y^{''}+4y=8tan^2(2x)
y
′
′
+
4
y
=
8
tan
2
(
2
x
)
y^{''}+4y^'=8
y
′
′
+
4
y
′
=
8
y^{'''}-4y^{''}+5y^'=40t-7-6e^{2t}
y
′
′
′
−
4
y
′
′
+
5
y
′
=
4
0
t
−
7
−
6
e
2
t
y^{''}+3y^'=-e^{2x}+e^{3x}
y
′
′
+
3
y
′
=
−
e
2
x
+
e
3
x
y^{''}+4y^'+5y=e^{-2x}cos(x)
y
′
′
+
4
y
′
+
5
y
=
e
−
2
x
cos
(
x
)
1
..
2345
2346
2347
2348
2349
..
2459