Q. Find the particular solution, y=f(x), to the differential equation dxdy=ycosx given f(23π)=−1 and then find f(2π).
Separate and Integrate: Step 1: Separate variables and integrate.We start by separating the variables in the differential equation dxdy=ycosx. Rearrange to get ydy=cosxdx.Integrate both sides:∫ydy=∫cosxdx,2y2=sinx+C,
Solve for y: Step 2: Solve for y.To find y, rearrange the equation:y2=2(sinx+C),y=±2(sinx+C),
Find C: Step 3: Use the initial condition to find C.Given f(23π)=−1, plug in x=23π and y=−1:(−1)2=2(sin(23π)+C),1=2(−1+C),1=−2+2C,2C=3,C=23,
Write Solution: Step 4: Write the particular solution.With C=23, the solution becomes:y=±2(sinx+23),Since f(23π)=−1, we choose the negative branch:y=−2(sinx+23),
Find f(2π): Step 5: Find f(2π).Plug in x=2π:f(2π)=−2(sin(2π)+23),f(2π)=−2(1+23),f(2π)=−2×2.5,f(2π)=−5,
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