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cluponka [151]
3 years ago
9

First-order linear differential equations

Mathematics
1 answer:
garik1379 [7]3 years ago
4 0

Answer:

(a) \frac{dy}{(2y+1)}=tdt (b) y=\frac{e^{t^2}+e^{2c}-1}{2}

Step-by-step explanation:

(1) We have given \frac{dy}{dt}+ycost=0

\frac{dy}{dt}=-ycost

\frac{dy}{y}=-costdt

Integrating both side

lny=-sint+c

y=e^{-sint}+e^{-c}

(2) \frac{dy}{dt}-2ty=t

\frac{dy}{dt}=2ty+t

\frac{dy}{dt}=t(2y+1)

\frac{dy}{(2y+1)}=tdt

On integrating both side  

\frac{ln(2y+1)}{2}=\frac{t^2}{2}+c

ln(2y+1)={t^2}+2c

2y+1=e^{t^2}+e^{2c}

y=\frac{e^{t^2}+e^{2c}-1}{2}

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