please explain in detailed steps the general solution of the following differential equation. given equation: y-3y'/x=x solution: C1x^4/4-x^3/3+C2 required : detailed steps in the solution of the above

Dear student,

Let y=e^mx

dy/dx=me^mx

2y/dx2=m2e^mx

Now substitute this and solve the equation......

y¹¹ - 3y¹/x = x

d/dx (y¹) - 3y¹/x = x             ..................1                    

put y¹ = z theneq 1 becomes

dz/dx - 3z/x = x               ..............2

now this is a linear differential equation withintegration factor = 1/x³

z/x³ = x/x³ dx

z/x³ = (1/x²)dx = -1/x + c₁

z = -x² + c₁x³ ...................3

now , dy/dx = z so

dy/dx = -x² + c₁x³

dy = (-x² + c₁x³)dx

 y  = -x³/3 + c₁x⁴/4 + c₂           .................5

this is the required equation

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