Question

Find the solution of the differential equation that satisfies the given initial condition. (du)/(dt) = (2t + sec^2 t)/(2u), u(0) = -1

Answer 1

Ah okay so in differential equations you usually want the top variable isolated. To do this, multiply by dt and 2u and you get
$2udu = 2t + { \sec(t) }^{2} dt$
Now just integrate both sides. The integral of 2u with respect to u is u². The integral of (2t + sec²(t) with respect to t is t² + ∫sec²(t)dt. The last part is just tan(x) because d/dt(tan(t)) is sec²(t) so just integrating gets us back. Now we have
${u}^{2} + c = {t}^{2} + \tan(t) + k$
Where c and k are arbitrary constants. Subtracting c from k and you get
${u}^{2} = {t}^{2} + \tan(t) + b$
Where b is another constant. To find b, just plug in u(0) = -1 where u is -1 and t is 0. This becomes
$1 = \tan(0) + b$
tan(0) is 0 so b = 1. Take the plus or minus square root on both sides and you finally get
$u = \: { {t}^{2} } + \tan(t) + 1$
But Brainly didn't let me do but juat remember there is a plus or minus square root on the left.