Verify that y=e^x+1 is a solution of the differential equation y''-y'=0.
Hint. Differentiate y twice, then substitute into the equation to check both sides match.
Since y=e^x+1, y'=e^x, and y''=e^x. Substituting, y''-y' = e^x-e^x = 0, which matches the right-hand side, so the given function is a solution.
✦ Verified: y''-y'=0-0=0
Verify that y=x^2+2x+C is a solution of the differential equation y'-2x-2=0.
Hint. Differentiate y once and substitute into the equation.
Since y=x^2+2x+C, y'=2x+2. Substituting, y'-2x-2 = (2x+2)-2x-2 = 0, so the given function is a solution.
✦ Verified: y'-2x-2=0
Verify that y=cos x+C is a solution of the differential equation y'+sin x=0.
Hint. Differentiate y once and substitute into the equation.
Since y=cosx+C, y'=-sinx. Substituting, y'+sinx = -sinx+sinx = 0, so the given function is a solution.
✦ Verified: y'+sin x=0
Verify that y=sqrt(1+x^2) is a solution of the differential equation y'=xy/(1+x^2).
Hint. Differentiate y once using the chain rule, then substitute both y and y' into the right-hand side.
Since y=sqrt(1+x^2), y'=x/sqrt(1+x^2). Substituting y into the right-hand side, xy/(1+x^2) = x.sqrt(1+x^2)/(1+x^2) = x/sqrt(1+x^2), which equals y', so the given function is a solution.
✦ Verified: y'=xy/(1+x^2)
Verify that y=Ax is a solution of the differential equation xy'=y (x is not 0).
Hint. Differentiate y once and substitute into the equation.
Since y=Ax, y'=A. Substituting, xy'=xA=Ax=y, so the given function is a solution.
✦ Verified: xy'=y
Verify that y=x sin x is a solution of the differential equation xy'=y+x.sqrt(x^2-y^2) (x is not 0 and x>y or x<-y).
Hint. Differentiate y once using the product rule, then substitute into both sides.
Since y=x sinx, y'=sinx+x cosx, so xy'=x sinx+x^2 cosx. Since y+x sqrt(x^2-y^2) = x sinx+x sqrt(x^2-x^2 sin^2x) = x sinx+x^2|cosx|, and cosx matches |cosx| on the relevant domain, both sides agree.
✦ Verified: xy'=y+x.sqrt(x^2-y^2)
Verify that xy=log y+C is a solution of the differential equation y'=y^2/(1-xy) (xy is not 1).
Hint. Differentiate the implicit relation xy=logy+C with respect to x, then solve for y' and compare.
Differentiating xy=logy+C implicitly gives y+xy'=y'/y, so y^2+xyy'=y', giving y'(1-xy)=y^2, so y'=y^2/(1-xy), matching the given equation exactly.
✦ Verified: y'=y^2/(1-xy)
Verify that y-cos y=x is a solution of the differential equation (y siny+cosy+x)y'=y.
Hint. Differentiate the implicit relation y-cosy=x with respect to x, then substitute into the given equation.
Differentiating y-cosy=x implicitly gives y'+y'siny=1, so y'(1+siny)=1. Multiplying the given equation's factor by this relation and simplifying using x=y-cosy confirms both sides match.
✦ Verified: (y sin y+cos y+x)y'=y
Verify that x+y=tan^-1 y is a solution of the differential equation y^2 y'+y^2+1=0.
Hint. Differentiate the implicit relation x+y=tan^-1y with respect to x, then solve for y'.
Differentiating x+y=tan^-1y implicitly gives 1+y'=y'/(1+y^2), so y'[1-1/(1+y^2)]=-1, giving y'.y^2/(1+y^2)=-1, so y^2y'=-(1+y^2), which rearranges to y^2y'+y^2+1=0 exactly.
✦ Verified: y^2y'+y^2+1=0
Verify that y=sqrt(a^2-x^2), x in (-a,a), is a solution of the differential equation x+y(dy/dx)=0 (y is not 0).
Hint. Differentiate y once using the chain rule, then substitute into the equation.
Since y=sqrt(a^2-x^2), y'=-x/sqrt(a^2-x^2)=-x/y. Substituting, x+y.y' = x+y.(-x/y) = x-x = 0, so the given function is a solution.
✦ Verified: x+y(dy/dx)=0
Choose the correct answer: the number of arbitrary constants in the general solution of a differential equation of fourth order are (A) 0 (B) 2 (C) 3 (D) 4
Hint. Recall the standard relationship between the order of a differential equation and the number of arbitrary constants in its general solution.
Since the general solution of an n-th order differential equation contains exactly n independent arbitrary constants, a fourth order equation has 4.
✦ (D) 4
Choose the correct answer: the number of arbitrary constants in the particular solution of a differential equation of third order are (A) 3 (B) 2 (C) 1 (D) 0
Hint. Recall that a particular solution is obtained from the general solution by assigning specific values to every arbitrary constant.
Since a particular solution fixes every arbitrary constant to a specific numeric value using given conditions, it contains no arbitrary constants at all, regardless of the order.
✦ (D) 0
Solutions written by the tuition.in editorial team and checked against lemh203.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.
Pair self-study with a tutor, a live course or a coaching centre.
