Solve the equation (2xy − 3)dx + (x 2 + 4y)dy = 0 given that y(2) = 5.??

1 Answer

x^2y+2y^2-3x-64=0x2y+2y23x64=0

Explanation:

Given DE

(2xy-3)dx+(x^2+4y)dy=0(2xy3)dx+(x2+4y)dy=0

Comparing above equation with the standard form of DE Mdx+Ndy=0Mdx+Ndy=0 we get

M=2xy-3\implies \frac{\partial M}{\partial y}=2xM=2xy3My=2x &

N=x^2+4y\implies \frac{\partial N}{\partial x}=2xN=x2+4yNx=2x

Since, \frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}My=Nx hence the given DE is an exact DE whose solution is given as

int_{y=const} Mdx+\int_{\text{free of x}}Ndy =Cy=constMdx+free of xNdy=C

\int (2xy-3)\ dx+\int (x^2+4y)\ dy=C

2y\int x\ dx-3\int dx+4\int y\ dy=C

x^2y-3x+2y^2=C

Now, applying initial condition, y(2)=5 by setting x=2 & y=5 in above solution we get

2^2(5)-3(2)+2(5)^2=C

C=64

x^2y-3x+2y^2=64

x^2y+2y^2-3x-64=0