Find the equation of a curve that goes through the point (0,-2) and satisfies th

Question

Find the equation of a curve that goes through the point (0,-2) and satisfies that the slope at any of its points is equal to 3 plus the y coordinate at that point.

Find the general solution of the equation xy'=sqrt(x^2-y^2)+y.

If possible please make the notation as simple as possible so that the steps in between arent difficult to follow. Thanks!!

Explanation / Answer

Slope = 3+y =>dy/dx = 3+y =>dy/dx-y = 3 Multiplying on both sides by e^-x =>e^-xdy/dx-e^-xy = 3e^-x =>d(ye^(-x))/dx = 3e^-x =>ye^-x = -3e^-x + C =>y = -3+Ce^x The curve passes through (0,-2) =>-2 = -3+C =>C = 1 =>y = e^x-3......Required curve.

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