If `int(u(dv)/(dx))dx=uv-intwdx,` then `w=`

int(logx)^(2)dx=ux+c, then u=

If int(sinx)/(cos^(5)x)dx=intu^(3)du , then u=

int e ^ (- (v) / (u)) dv

If u and v are two differentiable functions of x and int uv dx=phi(x)-int[(du)/(dx)intv dx]dx , then the value of phi(x) is-

int(u(v(du)/(dx)-u(dv)/(dx)))/(v^(3))dx=

Consider the homogeneous differential equation (dy)/(dx)=f((y)/(x)) By putting (y)/(x)=v , the equation reduces to (dv)/(f(v)-v)=(dx)/(x) . Integrating we get int(dv)/(f(v)-v)=int(dx)/(x)+c Solution of xy(dy)/(dx)-y^(2)=(x+y)^(2)e^((-y)/(x)) is

Consider the homogeneous differential equation (dy)/(dx)=f((y)/(x)) By putting (y)/(x)=v , the equation reduces to (dv)/(f(v)-v)=(dx)/(x) . Integrating we get int(dv)/(f(v)-v)=int(dx)/(x)+c Solution of (x+ycos""(y)/(x))dx=xcos""(y)/(x)dy is

Consider the homogeneous differential equation (dy)/(dx)=f((y)/(x)) By putting (y)/(x)=v , the equation reduces to (dv)/(f(v)-v)=(dx)/(x) . Integrating we get int(dv)/(f(v)-v)=int(dx)/(x)+c Solution of x(x-y)dy+y^(2)dx=0 is-

Find the particular solution of v(dv)/(dx)=n^(2)x , given v=u when x=a

If u, v and w are functions of x, then show that d/(dx)(udotvdotw)=(d u)/(dx)vdotw+udot(d v)/(dx)dotw+udotv(d w)/(dx) in two ways - first by repeated application of product rule, second by logarithmic differentiation.