Differential eq

[fiery juniper]

Please help, I missed the day we went over this so idk how to start it

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[wanton finch]

Note that $\dv{x}\frac{1}{3}y^3=y^2y'$

[undone acorn]

Omegabet_

[wanton finch]

hence, integrating both sides wrt x, gives $\int\dv{x}\frac{1}{3}y^3\dd{x}=\int\sin(9x)\dd{x}$

[undone acorn]

Omegabet_

[wanton finch]

which means $\frac{1}{3}y^3=-\frac{1}{9}\cos(9x)+C$

[undone acorn]

Omegabet_

[wanton finch]

alternatively, and perhaps more typical/handwavy, you can use the shorthand version of noting $y^2\dv{y}{x}=\sin(9x)\implies y^2\dd{y}=\sin(9x)\dd{x}\to\int y^2\dd{y}=\int\sin(9x)\dd{x}$

[undone acorn]

Omegabet_

[fiery juniper]

Could you explain how these are equal