First we solve the integral by separating the terms:
d y d x = x + 3 ⟹ ∫ d y = ∫ ( x + 3 ) d x y = ∫ x 1 / 2 d x + 3 ∫ d x = x 3 / 2 3 / 2 + 3 x + C y = 2 x 3 / 2 3 + 3 x + C \frac{dy}{dx}=\sqrt{x}+3
\\ \implies \intop dy=\int(\sqrt{x}+3)dx
\\y=\int\ x^{1/2}dx+3\int dx=\frac{x^{3/2}}{3/2}+3x+C
\\y=\frac{2x^{3/2}}{3}+3x+C d x d y = x + 3 ⟹ ∫ d y = ∫ ( x + 3 ) d x y = ∫ x 1/2 d x + 3 ∫ d x = 3/2 x 3/2 + 3 x + C y = 3 2 x 3/2 + 3 x + C
Then, we substitute the coordinates (x,y) to find C for the particular solution:
3 = 2 ( − 1 ) 3 / 2 3 + 3 ( − 1 ) + C ⟹ 3 = − 2 i 3 − 3 + C → C = 6 + 2 i 3 3=\frac{2(-1)^{3/2}}{3}+3(-1)+C
\\ \implies 3=-\frac{2i}{3}-3+C
\\ \to C=6+\frac{2i}{3} 3 = 3 2 ( − 1 ) 3/2 + 3 ( − 1 ) + C ⟹ 3 = − 3 2 i − 3 + C → C = 6 + 3 2 i
In conclusion:
General solution: y = 2 x 3 / 2 3 + 3 x + C Particular solution: y = 2 x 3 / 2 3 + 3 x + 6 + 2 i 3 \text{General solution: }
\\ y=\frac{2x^{3/2}}{3}+3x+C
\\ \text{Particular solution: }
\\ y=\frac{2x^{3/2}}{3}+3x+6+\frac{2i}{3} General solution: y = 3 2 x 3/2 + 3 x + C Particular solution: y = 3 2 x 3/2 + 3 x + 6 + 3 2 i
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