∠BDC=30, so ∠CBD + ∠BCD=180-30=150 ΔDBC is isosceles, so ∠CBD=∠BCD=half of 150=75 ∠ABC=∠ABD-∠CBD=155-75=80 ΔABC is isosceles, so ∠ABC=∠ACB=80 ∠BAC=180-80-80=20
The general solution of <span>y' + 2xy = (x^3) is y(x) = c1e^(-x^2)
The general solution of </span><span>ydx = (y(e^y) - 2x)dy is x = c1/(y(x))^2 + (e^y(x) ((y(x))^2 - 2y(x) + 2)/(y(x))^2 </span> The general solution of<span> (dP)/(dt) + 2tP = P + 4t - 2 is P(t) = c1e^(t - t^2) + 2</span>
