Q. (3y2+2)dxdy=1 and y(−1)=1.What is x when y=2 ?x=
Separate variables: First, we need to separate the variables to integrate both sides of the differential equation. So we get \frac{dy}{\(3\)y^{\(2\)}+\(2\)} = \frac{dx}{\(1\)}\.
Integrate both sides: Now, integrate both sides. The integral of \(\frac{dy}{3y^{2}+2} with respect to y is the integral of dx with respect to x.
Find constant of integration: The integral of 3y2+2dy is 31⋅arctan(2/3y)+C, and the integral of dx is x.
Solve for C: We need to find the constant of integration C using the initial condition y(−1)=1. So we plug in y=1 and x=−1 into the integrated equation: (−31)⋅arctan(321)+C=−1.
Particular solution: Solve for C. C=−1+(31)⋅arctan(23).
Find x when y=2: Now we have the particular solution. We need to find x when y=2. Plug y=2 into the integrated equation: (1/3)⋅arctan(2/2/3)+C=x.
Substitute C into equation: Substitute the value of C we found earlier into the equation: x=(31)∗arctan(322)−1+(31)∗arctan(23).
Calculate x: Now, calculate the value of x. x≈(31)∗arctan(2∗23)−1+(31)∗arctan(23).
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