Question

another practice

Answer 1

\(\large\color{#000000 }{ \displaystyle y'-3xy=y^8 }\)

Answer 2

\(\large\color{#000000 }{ \displaystyle y'y^{-8}-3xy^{-7}=1 }\)

\(\large\color{#000000 }{ \displaystyle v=y^{-7} }\)
\(\large\color{#000000 }{ \displaystyle (-1/7)v'=y'y^{-8} }\)

Answer 3

\(\large\color{#000000 }{ \displaystyle (-1/7)v'-3xv=1 }\)
\(\large\color{#000000 }{ \displaystyle v'+21xv=-7 }\)

\(\large\color{#000000 }{ \displaystyle e^{H(x)}=e^{\frac{21}{2}x^2} }\)

Answer 4

\(\large\color{#000000 }{ \displaystyle v'e^{(21/2)x^2}+21xe^{(21/2)x^2}v=-7e^{(21/2)x^2} }\)

\(\large\color{#000000 }{ \displaystyle ve^{(21/2)x^2}=-7\int e^{(21/2)x^2}dx }\)

Answer 5

maybe the example is bad, because I don't want to go into the series at the moment, but in essence,

\(\large\color{#000000 }{ \displaystyle v=\frac{-7\int e^{(21/2)x^2}dx}{e^{(21/2)x^2}} }\)

\(\large\color{#000000 }{ \displaystyle y^{-8}=\frac{-7\int e^{(21/2)x^2}dx}{e^{(21/2)x^2}} }\)

\(\large\color{#000000 }{ \displaystyle y=\frac{\sqrt[8]{e^{(21/2)x^2}}}{\sqrt[8]{-7\int e^{(21/2)x^2}dx}} }\)

Answer 6

oh, mistake

Answer 7

those are supposed to be not 8th roots, but powers of 8.

Answer 8

maybe the example is bad, because I don't want to go into the series at the moment, but in essence,

\(\large\color{#000000 }{ \displaystyle v=\frac{-7\int e^{(21/2)x^2}dx}{e^{(21/2)x^2}} }\)

\(\large\color{#000000 }{ \displaystyle y^{-8}=\frac{-7\int e^{(21/2)x^2}dx}{e^{(21/2)x^2}} }\)

\(\large\color{#000000 }{ \displaystyle y=\frac{\left\{e^{(21/2)x^2}\right\}^8}{\left\{-7\displaystyle \int e^{(21/2)x^2}dx\right\}^8} }\)

Answer 9

why did you change v=y^(-7) to v=y^(-8)

Answer 10

oh, well then powers of 7... 11:38 pm in my location :)

Answer 11

good night i guess, tnx

Answer 12

\[y^{-7}=f(x) \implies (y^{-7})^{\frac{-1}{7}}=(f(x))^{\frac{-1}{7}}=\frac{1}{(f(x))^{\frac{1}{7}}}\]

Answer 13

\[y^{-7}=f(x) \implies y=\frac{1}{(f(x))^{\frac{1}{7}}}\]

Answer 14

so 7 root not power

Answer 15

@fbi2015

Answer 16

ah you left quick :p