Question

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

Answer 1

Step-by-step 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

$\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1$

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

$u(x,y)=x^{3}+y^{4}-xy=1$