(i) The degree of the differential equation `(d^(2)y)/(dx^(2))+e^(dy//dx)=0` is…
(ii) The degree of the differential equation `sqrt(1+((dy)/(dx))^(2))="x is".......`
(iii) The number of arbitrary constant in the general solution of differential equation of order three is..
(iv) `(dy)/(dx)+(y)/(x log x)=(1)/(x)` is an equation of the type....
(v) General solution of the differential equation of the type is givven by...
(vi) The solution of the differential `(xdy)/(dx)+2y=x^(2)` is....
(vii) The solution of `(1+x^(2))(dy)/(dx)+2xy-4xy^(2)=0` is...
(viii) The solution of the differential equation `ydx+(x+y)dy=0` is ....
(ix) Genergal solution of `(dy)/dx)+y=sin x` is....
(x) The solution of differential equation cot y `dx=xdy ` is....
(xi) The integrating factor of `(dy)/(dx)+y=(1+y)/(x)` is.....

Explanation

(i) Given differential equation is
`" "(d^(2)y)/(dx^(2))+e^((dy)/(dx))=0`
Degree of this equation is not defined.
(ii) Given differential equation is `sqrt(1+((dy)/(dx))^(2))=x`
So, degree of this equation is two.
(iii) There are three arbitary constists.
(iv) Given differential equation is `(dy)/(dx)+(y)/(xlogx)=(1)/(x)`
The equation is of the type `(dy)/(dx)+Py=Q`
(v) Given differential equation is
`" "(dx)/(dy)+P_(1)x=Q_(1)`
The equation of the form `" "x*IF=intQ(IF)dy+C" "i.e., xe^(int(Pdy))=intQ{e^(intPdy)}dy+C`
(vi) Given differential equation is
`" "x(dy)/(dx)+2y=x^(2)rArr(dy)/(dx)+(2y)/(x)=x`
This equation of the form `(dy)/(dx)+Py=Q`.
`therefore" "IF=e^(int(2)/(x)dx)=e^(2logx)=x^(2)`
The general solution is
`" "yx^(2)=intx*x^(2)dx+C`
`rArr" "yx^(2)=(x^(4))/(4)+C`
`rArr" "y=(x^(2))/(4)+Cx^(-2)`
(vii) Given differential equations is
`" "(1+x^(2))(dy)/(dx)+2xy-4x^(2)=0`
`rArr" "(dy)/(dx)+(2xy)/(1+x^(2))-(4x^(2))/(1+x^(2))=0`
`rArr" "(dy)/(dx)+(2x)/(1+x^(2))y=(4x^(2))/(1+x^(2)`
`therefore" "IF=e^(int(2x)/(1+x^(2))dx)`
Put `" "1+x^(2)=trArr2xdx=dt`
`therefore " "IF=e^(int(dt)/(t))=e^(logt)=e^(log(1+x^(2)))=1+x^(2)`
The general solution is
`" "y*(1+x^(2))=int(1+x^(2))(4x^(2))/((1+x^(2)))dx+C`
`rArr" "(1+x^(2))y=int4x^(2)dx+C`
`rArr" "(1+x^(2))y+4(x^(3))/(3)+C`
`rArr" "y=(4x^(3))/(3(1+x^(2)))+C(1+x^(2))^(-1)`
Given differential equation is
`rArr" "ydx+(x+xy)dy=0`
`rArr " "ydx+x(1+y)dy=0 `
`rArr" "(dx)/(-x)=((1+y)/(y))dy`
`rArr" "int(1)/(x)dx=-int((1)/(y)+1)dy" "` [on integrating]
`rArr" "log(x)=-log(y)-y+logA`
`" "log(xy)+y=logA`
`rArr" " xye^(y)=A`
`rArr" "xy=Ae^(-y)`
(ix) Given differential equation is
`" "y*e^(x)=inte^(x)sinxdx+C`
Let `" "I=inte^(x)sinxdx" "...(i)`
`" "I=sinxe^(x)-intcosxe^(x)dx`
`" "=sinxe^(x)-cosxe^(x)+int(-sinx)e^(x)dx`
`" "2I=e^(x)(sinx-cosx)`
`" "I=(1)/(2)e^(x)(sinx-cosx)`
From Eq. (i)
`" "y*e^(x)=(x)/(2)(sinx-cosx)+C`
`rArr" "y=(1)/(2)(sinx-cosx)+C*e^(-x)`
(x) Given differential equation is
`" "cotydx=xdy`
`rArr" "(1)/(x)dx=tanydy`
On intergrating both sides, we get
`rArr" "int(1)/(x)dx=inttanydy`
`rArr" "log(x)=log(sec y)+logC`
`rArr" "log((x)/(secy))=logC`
`rArr" "(x)/(secy)=C`
`rArr" "x=Csecy`
(xi) Given differential equation is
`" "(dy)/(dx)+y=(1+y)/(x)`
`" "(dy)/(dx)+y=(1)/(x)+(y)/(x)`
`rArr" "(dy)/(dx)+y(1-(1)/(x))=(1)/(x)`
`therefore" "IF=e^(int(1-(1)/(x))dx)`
`" "=e^(x)*e^(-logx)=(e^(x))/(x)`