Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 9 Ex 9.2 page covers all 12 questions, solved step-by-step — differentiating a given explicit or implicit function (sometimes more than once) and substituting back into the differential equation to confirm that L.H.S. equals R.H.S.
Questions 1–10 all follow the same verification pattern, but the functions get progressively trickier — from a simple exponential in Q1 to implicit relations that need implicit differentiation in Q7, Q8 and Q9, where you differentiate both sides with respect to x and then use the original equation itself to simplify the result. Questions 11 and 12 are conceptual MCQs on a rule worth memorising: a general solution of an nth order differential equation has exactly n independent arbitrary constants, while any particular solution — obtained by fixing those constants — has none.
Differentiating y=e^x+1: y'=e^x, and differentiating again, y''=e^x.
Substituting: y''-y'=e^x-e^x=0 = R.H.S.
Differentiating y=x^2+2x+C: y'=2x+2.
Substituting: y'-2x-2=(2x+2)-2x-2=0 = R.H.S.
Differentiating y=\cos x+C: y'=-\sin x.
Substituting: y'+\sin x=-\sin x+\sin x=0 = R.H.S.
Differentiating y=\sqrt{1+x^2}=(1+x^2)^{1/2}: y'=\dfrac{1}{2}(1+x^2)^{-1/2}\cdot 2x=\dfrac{x}{\sqrt{1+x^2}}.
R.H.S.: \dfrac{xy}{1+x^2}=\dfrac{x\sqrt{1+x^2}}{1+x^2}=\dfrac{x}{\sqrt{1+x^2}}, which equals y'.
Differentiating y=Ax: y'=A.
L.H.S.: xy'=xA=Ax=y = R.H.S.
Differentiating y=x\sin x using the product rule: y'=\sin x+x\cos x.
L.H.S.: xy'=x\sin x+x^2\cos x.
Since y=x\sin x, we get x^2-y^2=x^2(1-\sin^2 x)=x^2\cos^2 x, so \sqrt{x^2-y^2}=|x\cos x|. Under the given domain restriction this simplifies to x\cos x, so R.H.S.=y+x\sqrt{x^2-y^2}=x\sin x+x\cdot x\cos x=x\sin x+x^2\cos x, matching L.H.S.
Differentiating xy=\log y+C implicitly with respect to x (product rule on the left): y+xy'=\dfrac{1}{y}y'.
Rearranging: y=y'\left(\dfrac{1}{y}-x\right)=y'\cdot\dfrac{1-xy}{y}.
So y'=\dfrac{y^2}{1-xy}, which matches the given equation.
Differentiating y-\cos y=x implicitly: y'+\sin y\cdot y'=1, so y'(1+\sin y)=1.
From the original equation, x=y-\cos y, so y\sin y+\cos y+x=y\sin y+\cos y+y-\cos y=y(\sin y+1).
Therefore L.H.S.=y(\sin y+1)\cdot y'=y(\sin y+1)\cdot\dfrac{1}{1+\sin y}=y = R.H.S.
Differentiating x+y=\tan^{-1}y implicitly: 1+y'=\dfrac{y'}{1+y^2}.
Rearranging: 1=y'\left(\dfrac{1}{1+y^2}-1\right)=y'\cdot\dfrac{1-(1+y^2)}{1+y^2}=\dfrac{-y^2y'}{1+y^2}.
So 1+y^2=-y^2y', i.e. y^2y'+y^2+1=0, matching the given equation.
Differentiating y=(a^2-x^2)^{1/2}: y'=\dfrac{-x}{\sqrt{a^2-x^2}}=\dfrac{-x}{y}.
Substituting: x+y\cdot y'=x+y\left(\dfrac{-x}{y}\right)=x-x=0.
The general solution of a differential equation contains as many independent arbitrary constants as the order of the equation.
A particular solution is obtained by assigning specific values to the arbitrary constants of the general solution, so it is free of arbitrary constants — regardless of the order of the equation.
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