To solve y' = e" + y – 1 we first rewrite the equation in the form dy – e + y – 1, and then (1 – e2* – y)dz + dy = 0. It follows that the equation is . If we choose H(x) = e== then we obtain e"(1- e – y)dx +e *dy = 0, which is .Hence, the function is an .We solve this last problem to obtain y = ce" + e²" + A where A=-2 A=2 A--3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1

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To solve
y' = e +y– 1
we first rewrite the equation in the form
dy - +y-1,
dz
and then
(1 – e – y)dz + dy = 0.
It follows that the equation is
If we choose
H(z) = e=
then we obtain
e "(1- e – y)dr +e*dy = 0,
which is
Hence, the function
µ(z)
is an
We solve this last problem to obtain
y = ce* +e + A
where
A=-2
A=2
A=-3
not exact
exact
full
A=-1
A=3
multiplicator
integrating factor
A=1
Transcribed Image Text:To solve y' = e +y– 1 we first rewrite the equation in the form dy - +y-1, dz and then (1 – e – y)dz + dy = 0. It follows that the equation is If we choose H(z) = e= then we obtain e "(1- e – y)dr +e*dy = 0, which is Hence, the function µ(z) is an We solve this last problem to obtain y = ce* +e + A where A=-2 A=2 A=-3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1
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