Question

given y>0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0

Answer 1

`(\mathrm dy)/(\mathrm dx)=\frac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx`

`\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx`

`\implies \frac12y^2=x^3+2x^2+C`

When
`x=1` you have
`y=√(10)`, so


`\frac12(√(10))^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2`

and so the particular solution to the ODE is


`\frac12y^2=x^3+2x^2+2`

Then when
`x=0`, you get


`\frac12y^2=0^3+2(0)^2+2=2`

`\implies y^2=4`

`\implies y=2`

where we omitted the negative root because it's given that
`y>0`.