Distribute Terms -5(x +2)
2 answers:
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Ah okay so in differential equations you usually want the top variable isolated. To do this, multiply by dt and 2u and you get
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
Where c and k are arbitrary constants. Subtracting c from k and you get
Where b is another constant. To find b, just plug in u(0) = -1 where u is -1 and t is 0. This becomes
tan(0) is 0 so b = 1. Take the plus or minus square root on both sides and you finally get
But Brainly didn't let me do but juat remember there is a plus or minus square root on the left.
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
Where c and k are arbitrary constants. Subtracting c from k and you get
Where b is another constant. To find b, just plug in u(0) = -1 where u is -1 and t is 0. This becomes
tan(0) is 0 so b = 1. Take the plus or minus square root on both sides and you finally get
But Brainly didn't let me do but juat remember there is a plus or minus square root on the left.