InterviewSolution
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InterviewSolution
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(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..... |
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Answer» (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)` |
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