f(y'',y',y) = 0 difererencialine lygtis. Kažkur darau klaidą

Kur darau klaidą ?

[tex]5{y}'^{2} + {y}''(1+y) = 0[/tex]

[tex]5{z}^{2} + {z}\frac{dz}{dy}(1+y) = 0[/tex]

[tex]5{z} + \frac{dz}{dy}(1+y) = 0[/tex]

[tex]\frac{5z}{dz} =- \frac{1+y}{dy}[/tex]

[tex]\int \frac{dz}{z} =-5 \int \frac{dy}{1+y}[/tex]

[tex]lnz =ln(1+y)^{-5} + lnC1[/tex]

[tex]z =\frac{C1}{(1+y)^{5}}[/tex]

[tex]y = -\frac{C1}{4}(1+y)^{-4} + C2[/tex]

CidoKur darau klaidą ?

[tex]5{y}'^{2} + {y}''(1+y) = 0[/tex]

[tex]5{z}^{2} + {z}\frac{dz}{dy}(1+y) = 0[/tex]

[tex]5{z} + \frac{dz}{dy}(1+y) = 0[/tex]

[tex]\frac{5z}{dz} =- \frac{1+y}{dy}[/tex]

[tex]\int \frac{dz}{z} =-5 \int \frac{dy}{1+y}[/tex]

[tex]lnz =ln(1+y)^{-5} + lnC1[/tex]

[tex]z =\frac{C1}{(1+y)^{5}}[/tex]

[tex]y = -\frac{C1}{4}(1+y)^{-4} + C2[/tex]

verbunai

CidoKur darau klaidą ?

[tex]5{y}'^{2} + {y}''(1+y) = 0[/tex]

[tex]5{z}^{2} + {z}\frac{dz}{dy}(1+y) = 0[/tex]

[tex]5{z} + \frac{dz}{dy}(1+y) = 0[/tex]

[tex]\frac{5z}{dz} =- \frac{1+y}{dy}[/tex]

[tex]\int \frac{dz}{z} =-5 \int \frac{dy}{1+y}[/tex]

[tex]lnz =ln(1+y)^{-5} + lnC1[/tex]

[tex]z =\frac{C1}{(1+y)^{5}}[/tex]

[tex]y = -\frac{C1}{4}(1+y)^{-4} + C2[/tex]

Darai esminę klaidą pamiršdamas, kad y=y(x).