Question

The differential equation whose solution represents the family \[x^{2}y=4e^{x}+c\], where c is an arbitrary constant, is

पर्याय

• \[x\frac{\mathrm{d}y}{\mathrm{d}x}+xy=0\]

• \[x^2\frac{\mathrm{d}y}{\mathrm{d}x}+\left(2x-xy\right)=0\]

• \[x\frac{\mathrm{d}y}{\mathrm{d}x}+\left(x-2\right)y=0\]

• \[x^2y\frac{\mathrm{d}y}{\mathrm{d}x}+2xy-4\mathrm{e}^x\]

Answer 1

\[x^2y\frac{\mathrm{d}y}{\mathrm{d}x}+2xy-4\mathrm{e}^x\]

Explanation:

\[x^2y=4\mathrm{e}^x+\mathrm{c}\]

Differentiating w.r.t.x, we get

\[2xy+x^2y\frac{\mathrm{d}y}{\mathrm{d}x}=4\mathrm{e}^x\]

\[\therefore\quad x^2y\frac{\mathrm{d}y}{\mathrm{d}x}+2xy-4\mathrm{e}^x=0\]