dydx=y+1 Solving the above given differential equation, yields the following general solution. y+1=ex+Cy=Cex−1⇒ Solution y=−1 at C=0Can I say that y=−1 is a singular solution?

Question

etsahalen5tt · Accepted Answer

Step 1y=−1  is not a singular solution because it is included in the general solution y=Cex−1 when C=0.Solutions are only called singular when they are not attainable from the general solution form. In fact, I believe all linear first order homogenous equations have no singular solutions.

Drahthaare89c · Accepted Answer

Step 1y=−1  is not a singular solution, because it does not pass by any point (x0, y0) such that the initial value problem{y′=y+1y(x0)=y0has more than one solution: in point of fact, the Cauchy problem{y′=y+1y(x0)=−1has only the one maximal solution y=−1, whatever x0 is.

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}+{1}\) Solving the above given differential equation,

Answered question

Answer & Explanation

New Questions in Calculus 2