Answer to Question #121098 in Differential Equations for Joseph Se

Question

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) = _____

Answer 1

(i) Given

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

Solving the differential equation,

Equation is linear in y and first order equation.

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

multiplying this factor on both sides of equation,

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

$e^{-t}(\frac{dy}{dx} - y) = 3te^{t}$

Now integration both sides with respect to x,

$e^{-t}y = \int 3te^{t}dt$

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

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

$e^{-t}y = 3e^{t}(t-1) + c$

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

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

$1 = -3 +c\implies c=4$

so equation will be

$y = 3e^{2t}(t-1) + 4 e^t$

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

integrating both sides

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

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

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

then $6 = 0 - 0 +c \implies c = 6$

then equation will be

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

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

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