du | ||
uy | −u2=y2ln(y) | |
dy |
du | u2+y2ln(y) | ||
= | |||
dy | uy |
y * y'' − (y')2 | |
= ln(y), żeby to lepiej zauważyć zapiszę lewą stronę w takiej formie: | |
y2 |
(y')' * y − y' * y' | |
= ln(y), widać, że lewa strona jest pochodną pewnego ilorazu, | |
y2 |
y' | ||
( | )' = ln(y) | |
y |
du | u | yln(y) | |||
= | + | ||||
dy | y | u |
du | u | yln(y) | |||
= | + | | *(2u) | |||
dy | y | u |
du | 2u2 | |||
2u | − | =2yln(y) | ||
dy | y |
dw | du | ||
=2u | |||
dy | dy |
dw | 2w | ||
− | =2yln(y) | ||
dy | y |
dw | 2w | ||
− | =0 | ||
dy | y |
dw | 2w | ||
= | |||
dy | y |
dw | 2 | ||
= | dy | ||
w | y |
dw | 2w | ||
− | =2yln(y) | ||
dy | y |
2 | ||
C'(y)y2+2yC(y)− | C(y)y2 = 2yln(y) | |
y |
2ln(y) | ||
C'(y) = | ||
y |
dy | |
= ±y√ln2(y)+C1 | |
dt |
dy | |
= ±dt | |
y√ln2(y)+C1 |
dy | |
= ±dt | |
y√ln2(y)+C1 |
dy | |
= dz | |
y |
dz | ||
∫ | = ±t+C2 | |
√z2+C1 |
x2 − C1 | ||
z = | ||
2x |
2x*2x−2(x2 − C1) | ||
dz = | dx | |
4x2 |
2x2+2C1 | ||
dz = | dx | |
4x2 |
x2+C1 | ||
dz = | dx | |
2x2 |
2x2−(x2−C1) | ||
x−z = | ||
2x |
x2+C1 | ||
x−z = | ||
2x |
2x | x2+C1 | dx | ||||
∫ | * | dx =∫ | ||||
x2+C1 | 2x2 | x |
(C22e±2t−C1)e±t | ||
z = | ||
2C2 |
(C22e±2t−C1)e±t | ||
ln(y) = | ||
2C2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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