Question
The solution of dydx+1=ex+y is
Answer: Option A
:
A
x+y=z⇒1+dydx=dzdxdydx+1=ex+y⇒dydx=e−zdz=dx⇒∫e−zdz=∫dx⇒−e−z=x+c⇒e−(x+y)+x+c=0.
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A
x+y=z⇒1+dydx=dzdxdydx+1=ex+y⇒dydx=e−zdz=dx⇒∫e−zdz=∫dx⇒−e−z=x+c⇒e−(x+y)+x+c=0.
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