Question

Solve the differential equation:

Find the differential equation of family of curves y = e^x (ax + bx²), where A and B are arbitrary constants.

Answer 1

y = e^x (ax + bx²)

∴ y = xe^x (a + bx)

∴ `y/(xe^x) = a+bx`

Differentiating w.r.t. x, we get

`(xe^x dy/dx -(y (e^x + xe^x)))/(x^2(e^x)^2) = b`

∴  `(xdy/dx -y -xy) /(x^2 e^x)  = b`

Again, differentiating w.r.t. x, we get

`(*x^2e^x(dy/dx + x(d^2y)/dx^2 - dy/dx -y - x dy/dx) - (xdy/dx - y -xy)(x^2e^x + 2xe^x))/(x^2e^x)^2 = 0`

∴ `xe^x[x(x (d^2y)/dx^2 -y -x dy/dx)-(xdy/dx -y -xy)(x+2)] = 0`

∴`x^2(d^2y)/dx^2 -xy -x^2 dy/dx - (x^2dy/dx -xy - x^2y + 2x dy/dx -2y - 2xy ) = 0`

∴ `x^2(d^2y)/dx^2 -xy -x^2 dy/dx -x^2 dy/dx + xy + x^2 y- 2x dy/dx + 2y + 2xy = 0`

∴`x^2 (d^2y)/dx^2 - 2x^2 dy/dx - 2x dy /dx + x^2 y + 2y +2xy =0`