# Find the implicit solution of the following initial value problem. (2y-3)dy = (6x^2 + 4x - 5)dx  ,         y(1) = 2

You need to integrate both sides such that:

`int (2y-3)dy = int (6x^2 + 4x - 5)dx =gt int 2y dy - int 3dy =`

`= int 6x^2 dx+ int 4x dx- int 5 dx =gt`

`=gt 2y^2/2 - 3y + c = 6x^3/3 + 4x^2/2 - 5x + c`

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You need to integrate both sides such that:

`int (2y-3)dy = int (6x^2 + 4x - 5)dx =gt int 2y dy - int 3dy =`

`= int 6x^2 dx+ int 4x dx- int 5 dx =gt`

`=gt 2y^2/2 - 3y + c = 6x^3/3 + 4x^2/2 - 5x + c`

`y^2 - 3y + c = 2x^3 + 2x^2 - 5x + c`

Since `y(1) = 2 =gt 2^2 - 3*2 + c = 2*1^3 + 2*1^2 - 5*1 + c`

`4 - 6 + c = 2 + 2 - 5 + c =gt -2 + 5 - 4 = c =gt c = -1`

The implicit solution to the given differential equation is `y^2 - 3y - 2x^3 - 2x^2 +5x - 1 = 0.`

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