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Calculus 2 help!?

Science & Mathematics by Anonymous 2018-06-27 21:36:50

Social Science

Calculus 2 help!?

5 answers

Find the general solution of y′=4(y−20).

  • Anonymous

    (1/4) ∫ dy / ( y - 20 ) = ∫ dx (1/4) log ( y - 20 ) = x + C log ( y - 20 ) = 4x + K y - 20 = e^(4x + K) y = e^K e^(4x) + 20 y = A e^(4x) + 20 y = A e^(4x) + 20

  • Anonymous

    dy/dx = 4y - 80 dx/dy = 1 / (4y - 80) x = 0.25*ln(4y - 80) + C 4(x - C) = ln(4y - 80) e^(4x - 4C) = 4y - 80 y = 0.25e^(4x - 4C) + 20

  • Anonymous

    dy/dx = 4(y - 20) dy/dx = 4y - 80 dy/(4y - 80) = dx ∫ dy/(4y - 80) = ∫ dx ln|4y - 80|/4 = x + c

  • Anonymous

    y' = 4 * (y - 20) dy/dx = 4 * (y - 20) dy / (y - 20) = 4 * dx Integrate ln|y - 20| = 4x + C y - 20 = e^(4x + C) y = 20 + C * e^(4x) y = 30 when x = 0 30 = 20 + C * e^(4 * 0) 10 = C * e^(0) 10 = C * 1 10 = C y = 20 + 10 * e^(4x) Can you do the next one?

  • Anonymous

    dy/(y-4)=4dt integrate both sides ---> y=Ce^(4t)+20 2/ y(0)=30--->C=... 3/ y(0)=8 --->C=...

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