Rakesh tried to solve the differential equation dxdy=2y⋅cos(x). This is his work:dxdy=2y⋅cos(x)Step 1: ∫y1dy=∫2cos(x)dxStep 2: ln∣y∣=2sin(x)Step 3: eln∣y∣=e2sin(x)Step 4: ∣y∣=e2sin(x)Step 5: y=±e2sin(x)+CIs Rakesh's work correct? If not, what is his mistake?Choose 1 answer:(A) Rakesh's work is correct.(B) Step 1 is incorrect. The separation of variables wasn't done correctly.(C) Step 2 is incorrect. The right-hand side of the equation should be 2sin(x)+C.(D) Step 4 is incorrect. eln∣y∣ isn't equal to ∣y∣.
Q. Rakesh tried to solve the differential equation dxdy=2y⋅cos(x). This is his work:dxdy=2y⋅cos(x)Step 1: ∫y1dy=∫2cos(x)dxStep 2: ln∣y∣=2sin(x)Step 3: eln∣y∣=e2sin(x)Step 4: ∣y∣=e2sin(x)Step 5: y=±e2sin(x)+CIs Rakesh's work correct? If not, what is his mistake?Choose 1 answer:(A) Rakesh's work is correct.(B) Step 1 is incorrect. The separation of variables wasn't done correctly.(C) Step 2 is incorrect. The right-hand side of the equation should be 2sin(x)+C.(D) Step 4 is incorrect. eln∣y∣ isn't equal to ∣y∣.
Separate variables: Separate variables.Rakesh starts by separating the variables y and x.ydy=2cos(x)dxThis is the correct method for separating variables in a differential equation.
Integrate both sides: Integrate both sides.Rakesh integrates both sides of the equation.∫(y1)dy=∫2cos(x)dxln∣y∣=2sin(x)+CRakesh forgot to include the constant of integration on the right-hand side of the equation.