Find the general solution: dy/dx+2y=sin x.
Hint. This is linear with P=2, Q=sinx; find the integrating factor e^(2x) first.
The integrating factor is e^(integral of 2 dx)=e^(2x); multiplying through and integrating y.e^(2x)=integral of e^(2x)sinx dx (a standard reduction integral) gives the general solution.
✦ y=(2sin x-cos x)/5+Ce^(-2x)
Find the general solution: dy/dx+3y=e^(-2x).
Hint. This is linear with P=3, Q=e^(-2x); find the integrating factor e^(3x) first.
The integrating factor is e^(3x); multiplying through and integrating y.e^(3x)=integral of e^(3x)e^(-2x)dx=integral of e^x dx=e^x+C gives the general solution after dividing by e^(3x).
✦ y=e^(-2x)+Ce^(-3x)
Find the general solution: dy/dx+y/x=x^2.
Hint. This is linear with P=1/x, Q=x^2; find the integrating factor x first.
The integrating factor is e^(integral of dx/x)=x; multiplying through and integrating y.x=integral of x.x^2 dx=x^4/4+C gives the general solution after dividing by x.
✦ y=x^3/4+C/x
Find the general solution: dy/dx+(sec x)y=tan x, 0<=x<pi/2.
Hint. This is linear with P=secx, Q=tanx; find the integrating factor secx+tanx first.
The integrating factor is e^(integral of secx dx)=secx+tanx (the standard log|secx+tanx| antiderivative exponentiated); multiplying through and integrating gives the general solution.
✦ y(sec x+tan x)=sec x+tan x-x+C
Find the general solution: cos^2 x (dy/dx)+y=tan x, 0<=x<pi/2.
Hint. Divide through by cos^2x first to put the equation into standard linear form dy/dx+Py=Q.
Dividing by cos^2x gives dy/dx+y sec^2x=tanx sec^2x; the integrating factor is e^(integral of sec^2x dx)=e^(tanx), and integrating y.e^(tanx)=integral of tanx sec^2x e^(tanx) dx (via t=tanx) gives the general solution.
✦ y=tan x-1+Ce^(-tan x)
Find the general solution: x(dy/dx)+2y=x^2 log x.
Hint. Divide through by x first to put the equation into standard linear form; the integrating factor is x^2.
Dividing by x gives dy/dx+2y/x=x logx; the integrating factor is e^(integral of 2dx/x)=x^2, and integrating y.x^2=integral of x^3 logx dx (by parts) gives the general solution.
✦ y=(x^2/4)(4log x-1)/4+C/x^2, i.e. y.x^2 = (x^4/16)(4log x-1)+C
Find the general solution: x log x (dy/dx)+y=(2/x)log x.
Hint. Divide through by x logx first; the integrating factor turns out to be log x itself.
Dividing by x logx gives dy/dx+y/(x logx)=2/x^2; the integrating factor is e^(integral of dx/(x logx))=logx (via t=logx), and integrating y.logx=integral of (2/x^2)logx dx (by parts) gives the general solution.
✦ y=C/log x-2/x-2/(x log x)
Find the general solution: (1+x^2)dy+2xy dx=cot x dx (x is not 0).
Hint. Divide through by (1+x^2) first; the integrating factor turns out to be (1+x^2) itself.
Dividing by (1+x^2) gives dy/dx+2xy/(1+x^2)=cotx/(1+x^2); the integrating factor is e^(integral of 2x/(1+x^2)dx)=1+x^2, and integrating y(1+x^2)=integral of cotx dx=log|sinx|+C gives the general solution.
✦ y(1+x^2)=log|sin x|+C
Find the general solution: x(dy/dx)+y-x+xy cot x=0 (x is not 0).
Hint. Divide through by x first, grouping the y terms — the integrating factor turns out to be x sin x.
Dividing by x gives dy/dx+y(1/x+cotx)=1; the integrating factor is e^(integral of (1/x+cotx)dx)=x sinx, and integrating y.x sinx=integral of x sinx dx (by parts) gives the general solution.
✦ y=1/x-cot x+C/(x sin x)
Find the general solution: (x+y)dy/dx=1.
Hint. Treat x as the dependent variable — rewritten as dx/dy-x=y, this is linear in x with integrating factor e^(-y).
Rewriting as dx/dy=x+y gives dx/dy-x=y, linear in x(y); the integrating factor is e^(-y), and integrating x.e^(-y)=integral of y.e^(-y)dy (by parts) gives the general solution.
✦ x=-y-1+Ce^y
Find the general solution: y dx+(x-y^2)dy=0.
Hint. Treat x as the dependent variable — this is linear in x(y) with integrating factor 1/y.
Rewriting as dx/dy+x/y=y gives a linear equation in x(y); the integrating factor is e^(integral of dy/y)=y, and integrating x.y=integral of y.y dy=y^3/3+C gives the general solution.
✦ x=y^2/3+C/y
Find the general solution: (x+3y^2)dy/dx=y (y>0).
Hint. Treat x as the dependent variable — rewritten as dx/dy-x/y=3y, this is linear in x with integrating factor 1/y.
Rewriting as dx/dy=(x+3y^2)/y gives dx/dy-x/y=3y, linear in x(y); the integrating factor is e^(-integral of dy/y)=1/y, and integrating x/y=integral of 3 dy=3y+C gives the general solution.
✦ x=3y^2+Cy
Find a particular solution: dy/dx+2y tan x=sin x; y=0 when x=pi/3.
Hint. This is linear with integrating factor sec^2x; integrate then apply the given condition.
Since the integrating factor is e^(integral of 2tanx dx)=sec^2x, integrating y.sec^2x=integral of sinx sec^2x dx (simplified via t=cosx) and applying y(pi/3)=0 fixes the constant.
✦ y=cos x-2cos^2 x (equivalently y=(1-2cos x)cos x)
Find a particular solution: (1+x^2)dy/dx+2xy=1/(1+x^2); y=0 when x=1.
Hint. Divide through by (1+x^2); the integrating factor turns out to be (1+x^2) again.
Dividing by (1+x^2) gives dy/dx+2xy/(1+x^2)=1/(1+x^2)^2; the integrating factor is 1+x^2, and integrating y(1+x^2)=integral of dx/(1+x^2)=tan^-1x+C, then applying y(1)=0, fixes the constant.
✦ y=(tan^-1 x-pi/4)/(1+x^2)
Find a particular solution: dy/dx-3y cot x=sin 2x; y=2 when x=pi/2.
Hint. This is linear with integrating factor cosec^3x (from e^(-3.integral of cotx dx)); apply the given condition after integrating.
Since the integrating factor is e^(-3 integral of cotx dx)=cosec^3x (i.e. dividing by sin^3x), integrating and simplifying using sin2x=2sinxcosx, then applying y(pi/2)=2, fixes the constant.
✦ y=4sin^3 x-2sin^2 x (equivalently y=(4sin x-2)sin^2 x)
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x,y) is equal to the sum of the coordinates of the point.
Hint. Translate the word problem into dy/dx=x+y, a linear equation with integrating factor e^(-x).
Since the condition translates to dy/dx-y=x, linear with integrating factor e^(-x), integrating y.e^(-x)=integral of x.e^(-x)dx (by parts) and applying the origin condition y(0)=0 fixes the constant.
✦ y=-x-1+e^x
Find the equation of a curve passing through the point (0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Hint. Translate the word problem into x+y-(dy/dx)=5, a linear equation with integrating factor e^(-x).
The condition translates to dy/dx=x+y-5, i.e. dy/dx-y=x-5, linear with integrating factor e^(-x); integrating y.e^(-x)=integral of (x-5)e^(-x)dx (by parts) gives y=(4-x)+Ce^x, and applying the point (0,2) fixes C=-2.
✦ y=4-x-2e^x
Choose the correct answer: the Integrating Factor of the differential equation x(dy/dx)-y=2x^2 is (A) e^(-x) (B) e^(-y) (C) 1/x (D) x
Hint. Divide through by x first to put the equation into standard linear form, then read off P(x).
Dividing by x gives dy/dx-y/x=2x, so P=-1/x; the integrating factor is e^(integral of -dx/x)=e^(-log x)=1/x.
✦ (C) 1/x
Choose the correct answer: the Integrating Factor of the differential equation (1-y^2)(dx/dy)+yx=ay (-1<y<1) is (A) 1/(y^2-1) (B) 1/sqrt(y^2-1) (C) 1/(1-y^2) (D) 1/sqrt(1-y^2)
Hint. Divide through by (1-y^2) first to put the equation into standard linear form in x(y), then read off P(y).
Dividing by (1-y^2) gives dx/dy+xy/(1-y^2)=ay/(1-y^2), so P=y/(1-y^2); the integrating factor is e^(integral of y/(1-y^2) dy)=e^(-(1/2)log(1-y^2))=1/sqrt(1-y^2).
✦ (D) 1/sqrt(1-y^2)
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.
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