Rx 2 y 2.

One should have 2yy ′ = (y2) ′ burned into one's memory. Now notice. x + (y2 / 2) ′ = √x2 + y2. The red observation tells us we could use the substitution v = y2. The blue observation tells us we can go further, since 2x = (x2) ′ (which is just our original observation a second time). Now: The area enclosed by the curves x 2 = y, y = x + 2 and x-axis is. View Solution. Q2. Find the area enclosed by the curve y = x 2 and the x-axis for interval x = 1 to x = 3. View Solution. Q3. The area enclosed between the curve y = 1 + x 2, the y-axis and the straight line y = 5 is given by. View Solution. Q4.If the circle x 2 + y 2 = a 2 intersect the hyperbola x y = c 2 in four points P (x 1, y 2), Q (x 2, y 2), R (x 3, y 3), S (x 4, y 4), then which of the following does not hold View Solution SolveIn the last video we had this second order linear homogeneous differential equation and we just tried it out the solution y is equal to e to the rx. And we figured out that if you try that out, that it works for particular r's. And those r's, we figured out in the last one, were minus 2 and minus 3.

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