Question #121098
1. Find the solution of the given initial value problem.
y prime - y = 3t(e^2t), y(0) = 1

y(t) = _____

2. Find the solution of the given initial value problem in explicit form.

y prime = (1-11x)y^2, y(0) = -1/6

Enclose numerators and denominators in parentheses. For example, (a-b) / (1+n).

y(x) = _____
1
Expert's answer
2020-06-11T20:04:49-0400

(i) Given

dydxy=3te2t\frac{dy}{dx} - y = 3te^{2t} with y(0)=1y(0) = 1


Solving the differential equation,

Equation is linear in y and first order equation.


Integral factor will be edt=ete^{ - \int dt} = e^{-t}


multiplying this factor on both sides of equation,

et(dydxy)=3te2tete^{-t}(\frac{dy}{dx} - y) = 3te^{2t}e^{-t}

et(dydxy)=3tete^{-t}(\frac{dy}{dx} - y) = 3te^{t}

Now integration both sides with respect to x,

ety=3tetdte^{-t}y = \int 3te^{t}dt

ety=3[tetetdt]e^{-t}y = 3[te^{t} - \int e^{t} dt]

ety=3[tetet]e^{-t}y = 3[te^{t} -e^{t}] +c+ c

ety=3et(t1)+ce^{-t}y = 3e^{t}(t-1) + c

so y=3e2t(t1)+cety = 3e^{2t}(t-1)+ce^{t}

applying condition, y(0)=1,y(0) = 1,


1=3+c    c=41 = -3 +c\implies c=4

so equation will be

y=3e2t(t1)+4ety = 3e^{2t}(t-1) + 4 e^t



(ii) dydx=(111x)y2\frac{dy}{dx} = (1-11x)y^2 with y(0)=16y(0) = \frac{-1}{6}

integrating both sides


dyy2=(111x)dx\int \frac{dy}{y^2} = \int (1-11x)dx


1y=x112x2+c\frac{-1}{y} = x - \frac{11}{2}x^2 + c


applying condition y(0)=16y(0) = \frac{-1}{6}

then 6=00+c    c=66 = 0 - 0 +c \implies c = 6

then equation will be


1y=x112x2+6\frac{-1}{y} = x - \frac{11}{2}x^2 + 6

1y=2x11x2+122\frac{-1}{y} = \frac{2x - 11x^2 + 12}{2}

then y=22x11x2+12{y} = \frac{-2}{2x - 11x^2 + 12}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS