Question

Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).

Answer 1

Explanation:

The given equation is

`(3x^(2)-y)dx+(4y^(3)-x)dy=0\\M(x,y)dx+N(x,y)dy=0`

As a check for exactness we have

`(\partial N)/(\partial x)=(\partial M)/(\partial y)\\\\\therefore (\partial N)/(\partial x)=(\partial (4y^(3)-x))/(\partial x) =-1\\\\(\partial M)/(\partial y)=(\partial (3x^(3)-y))/(\partial y) =-1\\\\\therefore (\partial N)/(\partial x)=(\partial M)/(\partial y)=-1`

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

`u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^(2)-y)\partial x+\phi (y,c)\\\\u(x,y)=x^(3)-xy+\phi (y,c)\\\\`

Similarly we have

`u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^(3)-x)\partial y+\phi (x,c)\\\\u(x,y)=y^(4)-xy+\phi (x,c)\\\\`

Comparing both the solutions we infer

`\phi (x,c)=x^(3)+c`

Hence the solution becomes

`u(x,y)=x^(3)+y^(4)-xy=c`

given boundary condition is that it passes through (1,1) hence

`1^(3)+1^(4)-1=c\\\\\therefore c=1`

thus solution is

`u(x,y)=x^(3)+y^(4)-xy=1`