Write as dy/dt=ky.
∫dy/y=k∫dt+c.
ln(y)=kt+c.
Therefore k=(ln(y)-c)/t. When y=5, t=1, ln(5)=k+c. k=ln(5)-c.
y=e^(kt+c)=e^(t(ln(5)-c)+c). dy/dt=(ln(5)-c)e^(t(ln(5)-c)+c).
When dy/dt=4, t=1: 4=5(ln(5)-c). ln(5)-c=0.8, so c=ln(5)-0.8.
Therefore k=0.8.
So y=e^(0.8t+ln(5)-0.8)=5e⁰˙⁸ᵗ(0.4493)=2.2466e⁰˙⁸ᵗ approx.