Find all solutions of the given differential equations and then find the particular solutions for which a point (x,y) is given:
dy/dx = √x + 3 ; (x,y) =(-1,3)
First we solve the integral by separating the terms:
dydx=x+3 ⟹ ∫dy=∫(x+3)dxy=∫ x1/2dx+3∫dx=x3/23/2+3x+Cy=2x3/23+3x+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+Cdxdy=x+3⟹∫dy=∫(x+3)dxy=∫ x1/2dx+3∫dx=3/2x3/2+3x+Cy=32x3/2+3x+C
Then, we substitute the coordinates (x,y) to find C for the particular solution:
3=2(−1)3/23+3(−1)+C ⟹ 3=−2i3−3+C→C=6+2i33=\frac{2(-1)^{3/2}}{3}+3(-1)+C \\ \implies 3=-\frac{2i}{3}-3+C \\ \to C=6+\frac{2i}{3}3=32(−1)3/2+3(−1)+C⟹3=−32i−3+C→C=6+32i
In conclusion:
General solution: y=2x3/23+3x+CParticular solution: y=2x3/23+3x+6+2i3\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=32x3/2+3x+CParticular solution: y=32x3/2+3x+6+32i
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