# Answer to Question #4314 in Discrete Mathematics for Melissa

Question #4314

Solve the equation

(x+2y)y'=1; y(0)=-1

(x+2y)y'=1; y(0)=-1

Expert's answer

Solve the equation (x+2y)y'=1

Equation type: y'=f((a1x+b1y+c1)/(a2x+b2y+c2) , delta=a1b2-a2b1

If

delta <>0 then

x=u+alfa,y=v+beta , where

- is the solution of system:a1alfa+b1 beta+c1=0a2alfa+b2 beta+c2=0

Find our delta=0If

delta=0 then u(x)=ax+by+c

So z=x+2y => dy/dx=0.5 dz/dx-0.5o.5 dz/dx-0.5=1/z => dz/dx=2/z+1=(2+z)/z => zdz/(z+2)=dxIntegrating the last equation we obtain z-ln|x+2y+2|=x+Cx+2y-2ln|x+2y+2|=x+CApplying initial conditions find that x+2y+2=0-2+2=0 which is incorrect because of lnSo where is no solution under given initial conditions

**Solution:**(x+2y)y'=1 => y'=1/(x+2y)Equation type: y'=f((a1x+b1y+c1)/(a2x+b2y+c2) , delta=a1b2-a2b1

If

delta <>0 then

x=u+alfa,y=v+beta , where

- is the solution of system:a1alfa+b1 beta+c1=0a2alfa+b2 beta+c2=0

Find our delta=0If

delta=0 then u(x)=ax+by+c

So z=x+2y => dy/dx=0.5 dz/dx-0.5o.5 dz/dx-0.5=1/z => dz/dx=2/z+1=(2+z)/z => zdz/(z+2)=dxIntegrating the last equation we obtain z-ln|x+2y+2|=x+Cx+2y-2ln|x+2y+2|=x+CApplying initial conditions find that x+2y+2=0-2+2=0 which is incorrect because of lnSo where is no solution under given initial conditions

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