Find the general solution: dy/dx=(1-cos x)/(1+cos x).
Hint. Rewrite the right-hand side using the half-angle identity, since it simplifies to tan^2(x/2).
Using the half-angle identities 1-cosx=2sin^2(x/2) and 1+cosx=2cos^2(x/2), the right-hand side simplifies to tan^2(x/2)=sec^2(x/2)-1, which integrates directly.
✦ y=2tan(x/2)-x+C
Find the general solution: dy/dx=sqrt(4-y^2), -2<y<2.
Hint. Separate the variables — this reduces to a standard sin^-1 integral on the y side.
Separating gives dy/sqrt(4-y^2)=dx, and integrating both sides using the standard sin^-1 form gives sin^-1(y/2)=x+C.
✦ y=2sin(x+C)
Find the general solution: dy/dx+y=1 (y is not 1).
Hint. Rewrite as dy/dx=1-y and separate the variables.
Separating gives dy/(1-y)=dx, and integrating both sides gives -log|1-y|=x+C1, which rearranges to the exponential form.
✦ y=1-Ce^(-x)
Find the general solution: sec^2 x tan y dx+sec^2 y tan x dy=0.
Hint. Divide through by tanx tany to separate the variables into a function of x alone and a function of y alone.
Dividing by tanx tany gives (sec^2x/tanx)dx+(sec^2y/tany)dy=0, and since each term integrates to a log of the tangent, this gives log|tanx|+log|tany|=C.
✦ tan x . tan y=C
Find the general solution: (e^x+e^-x)dy-(e^x-e^-x)dx=0.
Hint. Separate the variables — the right-hand side is exactly the derivative of the left-hand side's bracket.
Separating gives dy=[(e^x-e^-x)/(e^x+e^-x)]dx, and since the numerator is exactly the derivative of the denominator, integrating gives a log directly.
✦ y=log(e^x+e^(-x))+C
Find the general solution: dy/dx=(1+x^2)(1+y^2).
Hint. Separate the variables — both sides reduce to a standard tan^-1 integral.
Separating gives dy/(1+y^2)=(1+x^2)dx, and integrating both sides using the standard tan^-1 form on the left gives tan^-1y=x+x^3/3+C.
✦ y=tan(x+x^3/3+C)
Find the general solution: y log y dx-x dy=0.
Hint. Separate the variables — divide through by xy logy.
Separating gives dy/(y logy)=dx/x, and since the left side integrates to log|logy| (via t=logy), the result is log|logy|=log|x|+C, which exponentiates cleanly.
✦ log y=Cx, i.e. y=e^(Cx)
Find the general solution: x^5 (dy/dx)=-y^5.
Hint. Separate the variables — this reduces to integrating y^-5 and x^-5 separately.
Separating gives dy/y^5=-dx/x^5, and integrating both sides using the power rule gives -1/(4y^4)=1/(4x^4)+C1, which rearranges into a symmetric form in x and y.
✦ 1/x^4+1/y^4=C
Find the general solution: dy/dx=sin^-1 x.
Hint. This is directly integrable — no separation is needed since the right-hand side depends only on x.
Since the right-hand side depends only on x, integrating directly (by parts, with u=sin^-1x, dv=dx) gives the general solution.
✦ y=x.sin^-1 x+sqrt(1-x^2)+C
Find the general solution: e^x tan y dx+(1-e^x) sec^2 y dy=0.
Hint. Separate the variables — divide through by tan y (1-e^x).
Separating gives sec^2y/tany dy = -e^x/(1-e^x) dx, and integrating both sides (the right side via t=1-e^x) gives log|tany|=log|1-e^x|+C.
✦ tan y=C(1-e^x)
Find a particular solution: (x^3+x^2+x+1)(dy/dx)=2x^2+x; y=1 when x=0.
Hint. Factor the cubic on the left by grouping, then separate the variables and integrate using partial fractions.
Factoring by grouping, x^3+x^2+x+1=(x+1)(x^2+1), so separating and decomposing (2x^2+x)/[(x+1)(x^2+1)] into partial fractions, then integrating and applying y(0)=1, gives the constant.
✦ y=(1/2)log|x+1|+(3/4)log(x^2+1)-(1/2)tan^-1x+1
Find a particular solution: x(x^2-1)(dy/dx)=1; y=0 when x=2.
Hint. Separate the variables, then decompose 1/[x(x^2-1)] into partial fractions before integrating.
Separating gives dy=dx/[x(x-1)(x+1)], and decomposing into partial fractions then integrating, followed by applying y(2)=0, gives the particular solution.
✦ y=-log|x|+(1/2)log|x^2-1|+(1/2)log3-log2
Find a particular solution: cos(dy/dx)=a (a is a real number); y=1 when x=0.
Hint. Take the inverse cosine of both sides first — this reduces the equation to a direct integral.
Taking cos^-1 of both sides gives dy/dx=cos^-1a, a constant, so integrating directly gives a linear function of x; applying y(0)=1 fixes the constant.
✦ y=x.cos^-1 a+1
Find a particular solution: dy/dx=y tan x; y=1 when x=0.
Hint. Separate the variables — this reduces to a standard log-secant integral.
Separating gives dy/y=tanx dx, and integrating both sides gives log|y|=log|secx|+C; applying y(0)=1 gives C=0.
✦ y=sec x
Find the equation of a curve passing through the point (0,0) and whose differential equation is y'=e^x sin x.
Hint. Integrate the right-hand side directly (by parts, applied twice, since it is the classic e^x sin x reduction integral), then use the point to fix the constant.
Integrating e^x sinx by parts twice (or using the standard reduction result) gives y=(e^x/2)(sinx-cosx)+C; substituting the point (0,0) fixes C=1/2.
✦ y=(e^x sin x-e^x cos x)/2+1/2
For the differential equation xy(dy/dx)=(x+2)(y+2), find the solution curve passing through the point (1,-1).
Hint. Separate the variables, splitting each side into a polynomial-plus-fraction form before integrating.
Separating gives y/(y+2)dy=(x+2)/x dx; rewriting each fraction as 1 minus (or plus) a simpler term and integrating, then substituting the point (1,-1), fixes the constant.
✦ y-x+2=2log|x(y+2)|
Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Hint. Translate the word problem into the equation y.(dy/dx)=x, then separate the variables.
The condition translates to y(dy/dx)=x, so separating gives y dy=x dx; integrating both sides gives y^2-x^2=C, and substituting the point (0,-2) gives C=4.
✦ y^2-x^2=4
At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).
Hint. Translate the word problem into dy/dx=2(y+3)/(x+4), then separate the variables.
The condition translates to dy/dx=2(y+3)/(x+4), so separating and integrating gives log|y+3|=2log|x+4|+C, which exponentiates to a quadratic relation; substituting (-2,1) fixes the constant.
✦ y+3=(x+4)^2, i.e. y=(x+4)^2-3
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after t seconds.
Hint. Since dV/dt is constant, integrate directly to get V as a linear function of t, then convert to radius using V=(4/3)pi.r^3.
Since dV/dt=k, V=kt+C. At t=0, V=(4/3)pi(27)=36pi gives C=36pi. At t=3, V=(4/3)pi(216)=288pi gives k=84pi. So V(t)=84pi.t+36pi=(4/3)pi.r^3, which solves for r.
✦ r=(63t+27)^(1/3)
In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (log_e 2=0.6931).
Hint. Set up dP/dt=(r/100)P, solve, and use the doubling condition to solve for r.
Since dP/dt=(r/100)P, separating and integrating gives P=P0.e^(rt/100); doubling in 10 years means 2=e^(10r/100), so taking logs gives r/10=log_e2, solving for r.
✦ r=6.93% per year (approximately)
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5=1.648)?
Hint. Use the continuous-growth formula P=P0.e^(rt/100) with the given rate and time.
Since P=P0.e^(rt/100) with P0=1000, r=5, t=10, this gives P=1000.e^0.5, which evaluates using the given value of e^0.5.
✦ Rs 1648
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Hint. Set up dN/dt=kN, find k from the 10%-in-2-hours condition, then solve for the time to double.
Since dN/dt=kN, N=N0.e^(kt). The 10% increase in 2 hours gives e^(2k)=1.1, so k=(log1.1)/2. Doubling requires e^(kt)=2, so t=(log2)/k=2log2/log1.1.
✦ t=2 log 2 / log 1.1 hours (approximately 14.6 hours)
Choose the correct answer: the general solution of the differential equation dy/dx=e^(x+y) is (A) e^x+e^-y=C (B) e^x+e^y=C (C) e^-x+e^y=C (D) e^-x+e^-y=C
Hint. Separate the variables using e^(x+y)=e^x.e^y, then integrate.
Since e^(x+y)=e^x.e^y, separating gives e^-y dy=e^x dx, and integrating both sides gives -e^-y=e^x+C1, which rearranges into a symmetric form matching one option exactly.
✦ (A) e^x+e^(-y)=C
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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