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Find all solutions of the following linear differential equations.

Solution

2y + (x+1)y\' = 3x+3 Solve the linear equation ( dy(x))/( dx) (x+1)+2 y(x) = 3 x+3: Rewrite the equation: ( dy(x))/( dx)+(2 y(x))/(x+1) = 3 Let mu(x) = exp( integral 2/(x+1) dx) = (x+1)^2. Multiply both sides by mu(x): (x+1)^2 ( dy(x))/( dx)+(2 (x+1)) y(x) = 3 (x+1)^2 Substitute 2 (x+1) = ( d)/( dx)((x+1)^2): (x+1)^2 ( dy(x))/( dx)+( d)/( dx)((x+1)^2) y(x) = 3 (x+1)^2 Apply the reverse product rule f ( dg)/( dx)+( df)/( dx) g = ( d)/( dx)(f g) to the left-hand side: ( d)/( dx)((x+1)^2 y(x)) = 3 (x+1)^2 Integrate both sides with respect to x: integral ( d)/( dx)((x+1)^2 y(x)) dx = integral 3 (x+1)^2 dx Evaluate the integrals: (x+1)^2 y(x) = x^3+3 x^2+3 x+c_1, where c_1 is an arbitrary constant. Divide both sides by mu(x) = (x+1)^2: y(x) = (x^3+3 x^2+3 x+c_1)/(x+1)^2