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26 January, 13:18

Solve the differential equation t dy/dt + dy/dt = te ^y

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  1. 26 January, 13:37
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    The answer is

    t dy / dt + dy / dt = te ^ y

    I apply common factor 1/dt * (tdy+dy) = te^y

    I pass "dt" tdy+dy = (te^y) * dt

    I apply common factor (t+1) * dy = (te^y) * dt

    I pass "e^y" (1/e^y) dy = ((t+1) * t) * dt

    I apply integrals ∫ (1/e^y) dy = ∫ (t^2+t) * dt

    by property of integrals ∫ (1/e^y) dy = ∫ (e^-y) dy

    ∫ (e^-y) dy = ∫ (t^2+t) * dt

    I apply integrals

    -e^-y = (t^3/3) + (t^2/2) + C

    I apply natural logarithm to eliminate "e"

    -ln (e^-y) = -ln (t^3/3) + (t^2/2) + C

    y=ln (t^3/3) + (t^2/2) + C
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