Question

Solve the initial value problem `y'=ky^2\lnx`, with `y(1)=-1`?

My work:

`dy/dx=ky^2\lnx`
`dy/y^2=k\lnxdx`
`\inty^-2dy=\intk\lnxdx=k\int\lnxdx`

See answer below for the rest

Answer 1

Continued from `\color(crimson)(\text(above))`

Explanation:

`\color(indianred)(dy/dx=ky^2\lnx)`
`\color(palevioletred)(dy/y^2=k\lnxdx)`
`\color(lightcoral)(\inty^-2dy=\intk\lnxdx=k\int\lnxdx)`

the rest...
`y^-1/-1=k[\lnx(x)-\int1dx]`
`-1/y=kx\lnx-kx+C`
`y=-1/(kx\lnx-kx+C)`

Given `y(1)=-1`, then
`-1=-1/(k(1)\ln(1)-k(1)+C)`
`-1=-1/(k(0)-k+C)`
`-1/-1=-k+C`
`1=-k+C`, `C=1+k`

`\thereforey=-1/(kx\lnx-kx+(1+k))`