Question

Find the solution of the differential equation that satisfies the given initial condition. du dt 2u 2t + sec?(6), (0) = -5 U = Vz2+ 2 + tan(t) + 25

Answer 1

Solution du dt 2t + sec? (t) 2u u(0) = -5 ? First order separable Ordinary Differential Equation A first order separable ODE has the form of N(y) y' = M(x) du Substitute with u' (t) dt 2t + sec? (t) u' (t) 2u(t) Rewrite in the form of a first order separable ODE N (u(t)). u' (t) = M(t) u' (t) 2t + sec' (t) 2u(t) Multiply both sides by 2u(t) 2u(t) 2t + sec (t) u' (t) 2u(t) = 2u(t) u' (t) 2u(t) = 2t + sec' (t)

Rewrite in standard form 2u(t) u' (t) = 2t + sec? (t) N (u(t)). u' (t) = M(t) N (u(t)) = 2u(t), M(t) = 2t + sec? (t) → 2u(t) u' (t) = 2t + sec? (t) If N(y): y = M(x), y dy dx then /N()dy = /M(x)d M(x) dx, up to a constant I 2u (t)du(t) = 1 2t + sec? (t) dt 2t + sec? (t) dt Apply the Sum Rule : /(x) = g(x) dx = (x) dx + 3 (3) - 2nd + / sec* (t) dt = t? + tan(t) = 12 + tan(t) +01
$\\\mathrm{\Rightarrow \int \:2u\left(t\right)du\left(t\right)=2\cdot \int \:u\left(t\right)du\left(t\right)=2\cdot \frac{u\left(t\right)^{1+1}}{1+1}=u\left(t\right)^2=u\left(t\right)^2+c_2}\\ \\ \mathrm{u\left(t\right)^2+c_2=t^2+\tan \left(t\right)+c_1}\\ \\ \mathrm{u\left(t\right)^2=t^2+\tan \left(t\right)+c_1}\\ \\ \mathrm{\mathrm{Apply\:initial\:conditions}:}\\ \\ \mathrm{\mathrm{Plug\:in\:}t=0\mathrm{:}\quad u\left(0\right)^2=0^2+\tan \left(0\right)+c_1\quad \mathrm{,\:and\:use\:initial\:condition\:}u\left(0\right)=-5}\\ \\ \mathrm{\left(-5\right)^2=0^2+\tan \left(0\right)+c_1}\\ \\ \mathrm{c_1=25}\\ \\ \mathrm{\mathrm{For\:}u\left(t\right)=t^2+\tan \left(t\right)+c_1\mathrm{\quad plug\:in\quad }c_1=25}\\ \\ \mathrm{u\left(t\right)^2=t^2+\tan \left(t\right)+25}\\ \\ \mathrm{\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}}\\ \\$$\\\mathrm{u\left(t\right)=\sqrt{t^2+\tan \left(t\right)+25},\:u\left(t\right)=-\sqrt{t^2+\tan \left(t\right)+25}}\\ \\ \mathrm{\mathrm{Check\:solutions\:by\:applying\:initial\:conditions}}\\ \\ \mathrm{u\left(0\right)=-5}\\ \\ \mathrm{u\left(t\right)=\sqrt{t^2+\tan \left(t\right)+25}:\quad \:-5=\sqrt{0^2+\tan \left(0\right)+25}\mathrm{:\quad False}}\\ \\ \mathrm{u\left(t\right)=-\sqrt{t^2+\tan \left(t\right)+25}:\quad \:-5=-\sqrt{0^2+\tan \left(0\right)+25}\mathrm{:\quad True}}\\ \\ \mathrm{\mathrm{Therefore,\:the\:final\:solution\:for\:}u'\left(t\right)=\frac{2t+\sec ^2\left(t\right)}{2u\left(t\right)},\:\quad u\left(0\right)=-5\mathrm{\:is\:}}\\ \\ \mathrm{u\left(t\right)=-\sqrt{t^2+\tan \left(t\right)+25}}$