(1) f(x)′=3x+1+∫0xf(t)dt=3x+1+F(x)−F(0)f'_{(x)} = 3x + 1 + \int_0^x f_{(t)}dt = 3x + 1 + F_{(x)} - F_{(0)}f(x)′=3x+1+∫0xf(t)dt=3x+1+F(x)−F(0) これから f(x)′=3+f(x)f'_{(x)} = 3 + f_{(x)}f(x)′=3+f(x) (2) f(x)=y とおくとy′=3+yf_{(x)} = y \text{ とおくと} \quad y' = 3 + yf(x)=y とおくとy′=3+y ∫13+ydy=∫dxlog(3+y)=x+C1eC1=C2 とおくと\int \frac{1}{3+y} dy = \int dx \quad \log(3+y) = x + C_1 \quad e^{C_1} = C_2 \text{ とおくと}∫3+y1dy=∫dxlog(3+y)=x+C1eC1=C2 とおくと y=C2ex−3y = C_2 e^x - 3y=C2ex−3 (1) から f(0)=1f_{(0)} = 1f(0)=1 なので C2=4C_2 = 4C2=4 これから f(x)=4ex−3f_{(x)} = 4e^x - 3f(x)=4ex−3