Please Show all steps and work thank you! Find the solution of the given initial

Question

Please Show all steps and work thank you!

Find the solution of the given initial value problem dy / dt - y = 2te2t, y(0) = 1; ty? + 2y = t2 - t + 1, y(1) = 1 / 2, t > 0; dy / dx = 2 - ex / 3 + 2y, y(0) = 0; y? = ty(4 - y) / 1 + t, y(0) = y0; dy / dt = 2(1 + t)(1 + y2), y(0) = 0.

Explanation / Answer

a) dy/dt - y = 2te^(2t) R(x) = e^(integ -1 dt) = e^-t yR(x) = integ R(x) 2te^(2t) dt ye^-t = integ e^-t 2t e^(2t) dt ye^-t = integ 2t e^t dt = 2e^t(t - 1) + c y = 2(e^2t)(t-1) + ce^t y(0) = 1 => 1 = c - 2 => c = 3 hence y = 2(e^2t)(t-1) + 3e^t b) ty' + 2y = t^2 - t + 1 y' + 2/t y = t - 1 + 1/t R(x) = e^(integ 2/t dt) = e^( 2 ln t) = e^( ln t^2) = t^2 yR(x) = integ R(x) (t - 1 + 1/t) dt yt^2 = integ t^2(t - 1 + 1/t ) dt yt^2 = integ t^3 - t^2 + t dt yt^2 = t^4/4 - t^3/3 + t^2/2 + c y = t^2/4 - t/3 + 1/2 + ct^-2 y(1) = 1/2 1/2 = 1/4 - 1/3 + 1/2 + c c = 1/12 hence y = t^2/4 - t/3 + 1/2 + t^-2/12 c) dy/dx = (2 - e^x)/(3 + 2y) (3 + 2y) dy = (2 - e^x) dx integ (3+2y) dy = integ (2 -e^x) dx 3y + y^2 = 2x - e^x + c y(0) = 0 => 0 = - 1 + c => c = 1 hence 3y + y^2 = 2x - e^x + 1 d) dy/dt = ty(4-y)/(1 + t) integ dy/(y(4-y)) = integ t /(1+t) dt integ dy/y^2(4y^-1 -1) = integ [(1 + t) - 1]/(1+ t) dt integ y^-2 /(4y^-1 -1) dy = integ 1 - 1/(1+ t) dt let y^-1 = x -y^-2dy = dx integ -dx/(4x - 1) = integ 1 - 1/(1 +t) dt - 1/4 ln(4x - 1) = t - ln(1 + t) + c - ln(4x -1) = 4t - 4ln(1 + t) + c substitute x back - ln(4y^-1 - 1) = 4t - 4ln(1+t) + c - ln((4-y)/y) = 4t - 4 ln(1 + t) + c ln (y/(4-y)) = 4t - 4ln(1+ t) + c y(0) = yo ln(yo/4 - yo) = c hence, ln(y/4-y) = 4t - 4ln(1 + t) + ln(yo/4 - yo) e) dy/dt = 2(1+t)(1+y^2) integ dy/(1+y^2) = integ 2(1+t) dt arctan y = 2t + t^2 + c y(0) = 0 c = arctan 0 = 0 hence arctan y = 2t + t^2 => y = tan(2t + t^2)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.