NCERT Solutions · Class 12 Mathematics Differential Equations

First factorise the coefficient by grouping: \[x^{3}+x^{2}+x+1 = x^{2}(x+1)+(x+1)=(x+1)\left(x^{2}+1\right)\] So \[dy = \frac{2x^{2}+x}{(x+1)\left(x^{2}+1\right)}\,dx\] Resolve into partial fractions, using a linear numerator over the irreducible quadratic: \[\frac{2x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}\] \[2x^{2}+x = A\left(x^{2}+1\right)+(Bx+C)(x+1)\] Put \(\displaystyle x=-1\): \(\displaystyle 2-1 = 2A \Rightarrow A=\tfrac12\). Comparing coefficients of \(\displaystyle x^{2}\): \(\displaystyle 2 = A+B \Rightarrow B = \tfrac32\). Comparing constants: \(\displaystyle 0 = A + C \Rightarrow C = -\tfrac12\). \[y=\int\left[\frac{1}{2}\cdot\frac{1}{x+1}+\frac{3}{2}\cdot\frac{x}{x^{2}+1}-\frac{1}{2}\cdot\frac{1}{x^{2}+1}\right]dx\] \[y = \frac{1}{2}\log|x+1| + \frac{3}{4}\log\left(x^{2}+1\right)-\frac{1}{2}\tan^{-1}x + K\] (the \(\displaystyle \tfrac34\) comes from \(\displaystyle \tfrac32\cdot\tfrac12\), since \(\displaystyle \int \frac{x\,dx}{x^{2}+1}=\tfrac12\log\left(x^{2}+1\right)\)). Apply \(\displaystyle y=1\) when \(\displaystyle x=0\): \(\displaystyle 1 = \tfrac12\log 1 + \tfrac34\log 1 - \tfrac12\tan^{-1}0 + K = K\), so \(\displaystyle K=1\). Particular solution: \(\displaystyle y = \dfrac{1}{4}\left[2\log|x+1| + 3\log\left(x^{2}+1\right) - 2\tan^{-1}x\right] + 1\).

Separate and factorise the denominator: \[dy = \frac{dx}{x\left(x^{2}-1\right)} = \frac{dx}{x(x-1)(x+1)}\] Partial fractions with three distinct linear factors: \[\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\] \[1 = A(x-1)(x+1)+Bx(x+1)+Cx(x-1)\] Put \(\displaystyle x=0\): \(\displaystyle 1 = -A \Rightarrow A=-1\). Put \(\displaystyle x=1\): \(\displaystyle 1 = 2B \Rightarrow B=\tfrac12\). Put \(\displaystyle x=-1\): \(\displaystyle 1 = 2C \Rightarrow C=\tfrac12\). \[y = -\log|x| + \frac{1}{2}\log|x-1| + \frac{1}{2}\log|x+1| + K = \frac{1}{2}\log\left|\frac{x^{2}-1}{x^{2}}\right| + K\] Apply \(\displaystyle y=0\) when \(\displaystyle x=2\): \[0 = \frac{1}{2}\log\frac{3}{4}+K \quad\Rightarrow\quad K = \frac{1}{2}\log\frac{4}{3}\] Particular solution: \(\displaystyle y = \dfrac{1}{2}\log\left|\dfrac{4\left(x^{2}-1\right)}{3x^{2}}\right|\).

The equation is not separable as printed; first solve it for \(\displaystyle \dfrac{dy}{dx}\). Since \(\displaystyle \cos\left(\dfrac{dy}{dx}\right)=a\), a real derivative exists only if \(\displaystyle -1\le a\le 1\); taking the principal value of the inverse cosine, \[\frac{dy}{dx}=\cos^{-1}a\] The right-hand side is a constant, so integrating, \[y = x\cos^{-1}a + C\] Apply \(\displaystyle y=1\) when \(\displaystyle x=0\): \(\displaystyle 1 = 0 + C\), so \(\displaystyle C=1\). \[y = x\cos^{-1}a + 1\] Rearranging for the form usually quoted, for \(\displaystyle x\neq 0\): \[\frac{y-1}{x}=\cos^{-1}a \quad\Rightarrow\quad \cos\left(\frac{y-1}{x}\right)=a\] Particular solution: \(\displaystyle y = 1 + x\cos^{-1}a\), equivalently \(\displaystyle \cos\left(\dfrac{y-1}{x}\right)=a\), valid for \(\displaystyle |a|\le 1\).

Variables separable (assume \(\displaystyle y\neq 0\)): \[\frac{dy}{y}=\tan x\,dx\] Integrate, using \(\displaystyle \int \tan x\,dx = \log|\sec x|\): \[\log|y| = \log|\sec x| + \log C\] \[y = C\sec x\] Apply \(\displaystyle y=1\) when \(\displaystyle x=0\): \(\displaystyle 1 = C\sec 0 = C\), so \(\displaystyle C=1\). Particular solution: \(\displaystyle y = \sec x\), on the interval containing \(\displaystyle x=0\) where \(\displaystyle \cos x>0\), i.e. \(\displaystyle -\dfrac{\pi}{2}<x<\dfrac{\pi}{2}\).

Here \(\displaystyle \dfrac{dy}{dx}=e^{x}\sin x\), so \(\displaystyle y=\int e^{x}\sin x\,dx\). Evaluate by parts twice and solve for the integral. Let \(\displaystyle I=\int e^{x}\sin x\,dx\); taking \(\displaystyle \sin x\) as the first function, \[I = e^{x}\sin x - \int e^{x}\cos x\,dx = e^{x}\sin x - \left[e^{x}\cos x + \int e^{x}\sin x\,dx\right]\] \[I = e^{x}\sin x - e^{x}\cos x - I \quad\Rightarrow\quad 2I = e^{x}(\sin x-\cos x)\] \[y = \frac{e^{x}(\sin x-\cos x)}{2}+C\] The curve passes through \(\displaystyle (0,0)\): \[0 = \frac{e^{0}(0-1)}{2}+C = -\frac{1}{2}+C \quad\Rightarrow\quad C=\frac{1}{2}\] Equation of the curve: \(\displaystyle y = \dfrac{e^{x}(\sin x - \cos x)+1}{2}\), i.e. \(\displaystyle 2y-1 = e^{x}(\sin x - \cos x)\).

The equation is variables separable after dividing by \(\displaystyle x(y+2)\) (needs \(\displaystyle x\neq 0\), \(\displaystyle y\neq -2\)): \[\frac{y}{y+2}\,dy = \frac{x+2}{x}\,dx\] Neither side can be integrated directly; split each fraction first (this is the step to get right): \[\left(1-\frac{2}{y+2}\right)dy = \left(1+\frac{2}{x}\right)dx\] Integrating, \[y - 2\log|y+2| = x + 2\log|x| + C\] \[y - x = 2\log|x| + 2\log|y+2| + C = 2\log|x(y+2)| + C\] The curve passes through \(\displaystyle (1,-1)\), where \(\displaystyle x=1\) and \(\displaystyle y+2=1\): \[-1-1 = 2\log 1 + C \quad\Rightarrow\quad C = -2\] Solution curve: \(\displaystyle y - x + 2 = 2\log|x(y+2)|\), equivalently \(\displaystyle y-x+2 = \log\left[x^{2}(y+2)^{2}\right]\).

The slope of the tangent at \(\displaystyle (x,y)\) is \(\displaystyle \dfrac{dy}{dx}\). Translating the condition "slope \(\displaystyle \times\) \(\displaystyle y\)-coordinate \(\displaystyle =\) \(\displaystyle x\)-coordinate": \[y\,\frac{dy}{dx}=x\] Separate and integrate: \[y\,dy = x\,dx \quad\Rightarrow\quad \frac{y^{2}}{2}=\frac{x^{2}}{2}+C_{1}\] \[y^{2}-x^{2}=C,\qquad C = 2C_{1}\] The curve passes through \(\displaystyle (0,-2)\): \[(-2)^{2}-0^{2}=C \quad\Rightarrow\quad C=4\] Equation of the curve: \(\displaystyle y^{2}-x^{2}=4\), a rectangular hyperbola (the branch through \(\displaystyle (0,-2)\), i.e. \(\displaystyle y=-\sqrt{x^{2}+4}\)).

The slope of the segment joining \(\displaystyle (x,y)\) to \(\displaystyle (-4,-3)\) is \(\displaystyle \dfrac{y-(-3)}{x-(-4)}=\dfrac{y+3}{x+4}\). The condition "slope of tangent is twice this" gives the differential equation \[\frac{dy}{dx}=\frac{2(y+3)}{x+4}\] Separate (valid for \(\displaystyle y\neq -3\), \(\displaystyle x\neq -4\)) and integrate: \[\frac{dy}{y+3}=\frac{2\,dx}{x+4}\] \[\log|y+3| = 2\log|x+4| + \log A = \log\left[A(x+4)^{2}\right]\] \[y+3 = A(x+4)^{2}\] The curve passes through \(\displaystyle (-2,1)\): \[1+3 = A(-2+4)^{2}=4A \quad\Rightarrow\quad A=1\] Equation of the curve: \(\displaystyle y+3 = (x+4)^{2}\), a parabola.

Let \(\displaystyle r\) be the radius at time \(\displaystyle t\) and \(\displaystyle V=\dfrac{4}{3}\pi r^{3}\) the volume. "Volume changes at a constant rate" means \[\frac{dV}{dt}=k \quad\Rightarrow\quad \frac{d}{dt}\left(\frac{4}{3}\pi r^{3}\right)=k\] Integrating with respect to \(\displaystyle t\): \[\frac{4}{3}\pi r^{3}=kt + C\] Use \(\displaystyle r=3\) at \(\displaystyle t=0\): \(\displaystyle \dfrac{4}{3}\pi(27)=36\pi = C\). Use \(\displaystyle r=6\) at \(\displaystyle t=3\): \(\displaystyle \dfrac{4}{3}\pi(216)=288\pi = 3k+36\pi\), so \(\displaystyle 3k = 252\pi\) and \(\displaystyle k = 84\pi\). Hence \[\frac{4}{3}\pi r^{3}=84\pi t + 36\pi \quad\Rightarrow\quad r^{3}=63t+27\] Radius after \(\displaystyle t\) seconds: \(\displaystyle r = (63t+27)^{1/3}\) units (check: \(\displaystyle t=0\) gives \(\displaystyle 3\), \(\displaystyle t=3\) gives \(\displaystyle 216^{1/3}=6\)).

Let \(\displaystyle P\) be the principal at time \(\displaystyle t\) years. "Increases continuously at the rate of \(\displaystyle r\%\) per year" means \[\frac{dP}{dt}=\frac{r}{100}P\] Separate and integrate: \[\frac{dP}{P}=\frac{r}{100}\,dt \quad\Rightarrow\quad \log P = \frac{rt}{100}+C_{1} \quad\Rightarrow\quad P = P_{0}e^{rt/100}\] Here \(\displaystyle P_{0}=100\) (the value at \(\displaystyle t=0\)). The money doubles in $\displaystyle 10$ years, so \(\displaystyle P=200\) at \(\displaystyle t=10\): \[200 = 100\,e^{10r/100} \quad\Rightarrow\quad e^{r/10}=2 \quad\Rightarrow\quad \frac{r}{10}=\log_{e}2 = 0.6931\] \[r = 6.931\] Hence \(\displaystyle r = 6.931\%\) per year.

With \(\displaystyle P\) the principal at time \(\displaystyle t\) years, continuous growth at \(\displaystyle 5\%\) per year gives \[\frac{dP}{dt}=\frac{5}{100}P = \frac{P}{20}\] Separating and integrating, \[\frac{dP}{P}=\frac{dt}{20} \quad\Rightarrow\quad \log P = \frac{t}{20}+C_{1} \quad\Rightarrow\quad P = P_{0}e^{t/20}\] Given \(\displaystyle P_{0}=1000\) at \(\displaystyle t=0\), so \(\displaystyle P = 1000\,e^{t/20}\). At \(\displaystyle t=10\): \[P = 1000\,e^{10/20}=1000\,e^{0.5}=1000(1.648)\] After $\displaystyle 10$ years the deposit is worth Rs 1648.

Let \(\displaystyle y\) be the bacteria count at time \(\displaystyle t\) hours. "Rate of growth proportional to the number present" gives \[\frac{dy}{dt}=ky\] Separating and integrating, \[\frac{dy}{y}=k\,dt \quad\Rightarrow\quad \log y = kt + C_{1} \quad\Rightarrow\quad y = y_{0}e^{kt}\] with \(\displaystyle y_{0}=1{,}00{,}000\). A \(\displaystyle 10\%\) increase in $\displaystyle 2$ hours means \(\displaystyle y = 1{,}10{,}000\) at \(\displaystyle t=2\): \[1{,}10{,}000 = 1{,}00{,}000\,e^{2k} \quad\Rightarrow\quad e^{2k}=\frac{11}{10} \quad\Rightarrow\quad k = \frac{1}{2}\log\frac{11}{10}\] Now require \(\displaystyle y = 2{,}00{,}000\), i.e. \(\displaystyle e^{kt}=2\): \[kt = \log 2 \quad\Rightarrow\quad t = \frac{\log 2}{k}=\frac{2\log 2}{\log\dfrac{11}{10}}\] Numerically, \(\displaystyle t = \dfrac{2(0.6931)}{0.0953}\approx 14.5\). The count reaches \(\displaystyle 2{,}00{,}000\) in \(\displaystyle t = \dfrac{2\log 2}{\log(11/10)}\approx 14.5\) hours.

Use the law of exponents to factor the right-hand side into a function of \(\displaystyle x\) times a function of \(\displaystyle y\), which makes the equation separable: \[\frac{dy}{dx}=e^{x+y}=e^{x}e^{y}\] \[e^{-y}\,dy = e^{x}\,dx\] Integrating both sides, and noting \(\displaystyle \int e^{-y}dy = -e^{-y}\) (the sign here decides between the options): \[-e^{-y}=e^{x}+C_{1}\] \[e^{x}+e^{-y}=-C_{1}=\mathrm{C}\] Hence the correct option is (A) \(\displaystyle e^{x}+e^{-y}=\mathrm{C}\).