What is the solution of the equationln(dydx)+x=0?

[c] Consider the given differential equation ln(dydx)+x=0⇒ln(dydx)=−x ⇒dydx=e−x On separating the variables, we get dy=e−xdx, On integrating both sides, we get ∫dy=∫e−xdx ⇒y=e−x−1+C=−e−x+C⇒y+e−x=C