Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exactC=.

Answer with Step-by-step explanation:The given differential equation is [tex](2x+5y)dx+(5x-4y)dy=0[/tex]Now the above differential equation can be re-written as[tex]P(x,y)dx+Q(x,y)dy=0[/tex]Checking for exactness we should have[tex]\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}[/tex][tex]\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5[/tex][tex]\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5[/tex]As we see that the 2 values are equal thus we conclude that the given differential equation is exactThe solution of exact differential equation is given by[tex]u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)[/tex]The value of [tex]\phi (y)[/tex] can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)[tex]\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c[/tex]