If y = y(x) is the solution of the differential equation, eydydx−1=ex, such that y(0) = 0, then y(1) is equal to:

The given differential equation is eydydx-1=ex ⇒ey-xdydx-1=1Substitute y-x=tDifferentiate w.r.t. x both sidesdydx-1=dtdxThe given differential equation becomes etdtdx=1etdt=dx  ⇒∫etdt=∫dx⇒et=x+c ⇒ey-x=x+cSubstitute x=0  y=0 given y(0)=0 it gives c=1Hence the solution is ey-x=x+1put  x=1 to get the value of y1⇒ey-1=2 ⇒y-1=loge2 ⇒y=1+loge2=y(1)