NCERT Solutions · Class 12 Mathematics Differential Equations

To verify a proposed solution, differentiate it as many times as the equation requires and substitute into the equation; the function is a solution if the equation reduces to an identity.Given \(\displaystyle y=e^{x}+1\). Differentiating with respect to \(\displaystyle x\) (the derivative of the constant \(\displaystyle 1\) is \(\displaystyle 0\)): \[y^{\prime}=e^{x}.\] Differentiating once more: \[y^{\prime\prime}=e^{x}.\]Substituting in the left-hand side of the given equation: \[y^{\prime\prime}-y^{\prime}=e^{x}-e^{x}=0,\] which is the right-hand side.Hence \(\displaystyle y=e^{x}+1\) satisfies \(\displaystyle y^{\prime\prime}-y^{\prime}=0\), so it is a solution of the given differential equation.

Differentiate the given function and substitute into the equation.Given \(\displaystyle y=x^{2}+2x+\mathrm{C}\), where \(\displaystyle \mathrm{C}\) is an arbitrary constant. Differentiating with respect to \(\displaystyle x\), and using \(\displaystyle \dfrac{d}{dx}(\mathrm{C})=0\): \[y^{\prime}=2x+2.\] Note that the constant \(\displaystyle \mathrm{C}\) disappears on differentiation, which is why the same derivative serves for every member of the family.Substituting in the left-hand side: \[y^{\prime}-2x-2=(2x+2)-2x-2=0,\] which is the right-hand side.Hence \(\displaystyle y=x^{2}+2x+\mathrm{C}\) is a solution of \(\displaystyle y^{\prime}-2x-2=0\) for every value of \(\displaystyle \mathrm{C}\).

Differentiate the given function and substitute into the equation.Given \(\displaystyle y=\cos x+\mathrm{C}\). Using \(\displaystyle \dfrac{d}{dx}(\cos x)=-\sin x\) and \(\displaystyle \dfrac{d}{dx}(\mathrm{C})=0\): \[y^{\prime}=-\sin x.\] The sign here is the step to watch: the derivative of \(\displaystyle \cos x\) carries a minus sign.Substituting in the left-hand side: \[y^{\prime}+\sin x=-\sin x+\sin x=0,\] which is the right-hand side.Hence \(\displaystyle y=\cos x+\mathrm{C}\) is a solution of \(\displaystyle y^{\prime}+\sin x=0\) for every value of \(\displaystyle \mathrm{C}\).

Differentiate by the chain rule, then re-express the answer in terms of \(\displaystyle y\) itself.Given \(\displaystyle y=\sqrt{1+x^{2}}=(1+x^{2})^{1/2}\). By the chain rule, \[y^{\prime}=\tfrac{1}{2}\,(1+x^{2})^{-1/2}\cdot\frac{d}{dx}\left(1+x^{2}\right)=\tfrac{1}{2}\,(1+x^{2})^{-1/2}\cdot 2x=\frac{x}{\sqrt{1+x^{2}}}.\]The right-hand side of the equation is written in terms of \(\displaystyle y\), so multiply numerator and denominator by \(\displaystyle \sqrt{1+x^{2}}\): \[y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}\cdot\frac{\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}=\frac{x\sqrt{1+x^{2}}}{1+x^{2}}=\frac{x\,y}{1+x^{2}},\] since \(\displaystyle y=\sqrt{1+x^{2}}\).This is exactly the given differential equation, so \(\displaystyle y=\sqrt{1+x^{2}}\) is a solution of \(\displaystyle y^{\prime}=\dfrac{xy}{1+x^{2}}\).

Differentiate the given function and substitute into the equation.Given \(\displaystyle y=\mathrm{A}x\), where \(\displaystyle \mathrm{A}\) is an arbitrary constant. Differentiating with respect to \(\displaystyle x\): \[y^{\prime}=\mathrm{A}.\]Substituting in the left-hand side of the equation, for \(\displaystyle x\neq 0\): \[x\,y^{\prime}=x\cdot \mathrm{A}=\mathrm{A}x=y,\] which is the right-hand side.Hence \(\displaystyle y=\mathrm{A}x\) is a solution of \(\displaystyle xy^{\prime}=y\) \(\displaystyle (x\neq 0)\) for every value of \(\displaystyle \mathrm{A}\); geometrically this is the family of straight lines through the origin.

Differentiate by the product rule and then simplify the surd on the right-hand side using the Pythagorean identity.Given \(\displaystyle y=x\sin x\). By the product rule, \[y^{\prime}=\sin x+x\cos x.\]Left-hand side of the equation: \[x\,y^{\prime}=x\sin x+x^{2}\cos x=y+x^{2}\cos x,\qquad\text{since } y=x\sin x.\]Right-hand side: with \(\displaystyle y=x\sin x\), \[x^{2}-y^{2}=x^{2}-x^{2}\sin^{2}x=x^{2}\left(1-\sin^{2}x\right)=x^{2}\cos^{2}x,\] using \(\displaystyle \sin^{2}x+\cos^{2}x=1\). Hence \[\sqrt{x^{2}-y^{2}}=\sqrt{x^{2}\cos^{2}x}=\left|x\cos x\right|.\] This is the step most easily got wrong: a square root is non-negative, so it is \(\displaystyle |x\cos x|\), not automatically \(\displaystyle x\cos x\). Taking the root as \(\displaystyle x\cos x\) is legitimate precisely where \(\displaystyle x\cos x\geq 0\), and there \[y+x\sqrt{x^{2}-y^{2}}=y+x\left(x\cos x\right)=y+x^{2}\cos x.\] The stated conditions \(\displaystyle x\neq 0\) and \(\displaystyle x>y\) or \(\displaystyle x<-y\) guarantee \(\displaystyle x^{2}>y^{2}\), so the surd is real.Both sides equal \(\displaystyle y+x^{2}\cos x\), so \(\displaystyle y=x\sin x\) is a solution of \(\displaystyle xy^{\prime}=y+x\sqrt{x^{2}-y^{2}}\) on the intervals where \(\displaystyle x\cos x\geq 0\).

The function is given implicitly, so differentiate the relation as it stands, treating \(\displaystyle y\) as a function of \(\displaystyle x\) (implicit differentiation), and then solve the result for \(\displaystyle y^{\prime}\).Given \(\displaystyle xy=\log y+\mathrm{C}\). Differentiating both sides with respect to \(\displaystyle x\), using the product rule on the left and the chain rule on \(\displaystyle \log y\): \[y+x\,y^{\prime}=\frac{1}{y}\,y^{\prime}.\]Multiply throughout by \(\displaystyle y\) (which is legitimate since \(\displaystyle \log y\) exists only for \(\displaystyle y>0\), so \(\displaystyle y\neq 0\)): \[y^{2}+xy\,y^{\prime}=y^{\prime}.\]Collect the \(\displaystyle y^{\prime}\) terms on one side: \[y^{2}=y^{\prime}-xy\,y^{\prime}=y^{\prime}\left(1-xy\right).\]Since \(\displaystyle xy\neq 1\), we may divide by \(\displaystyle 1-xy\): \[y^{\prime}=\frac{y^{2}}{1-xy}.\]This is precisely the given differential equation, so \(\displaystyle xy=\log y+\mathrm{C}\) is an (implicit) solution of \(\displaystyle y^{\prime}=\dfrac{y^{2}}{1-xy}\), \(\displaystyle xy\neq 1\).

Differentiate the implicit relation, find \(\displaystyle y^{\prime}\), and then use the relation itself to eliminate \(\displaystyle x\) from the left-hand side.Given \(\displaystyle y-\cos y=x\). Differentiating both sides with respect to \(\displaystyle x\), treating \(\displaystyle y\) as a function of \(\displaystyle x\): \[y^{\prime}-\left(-\sin y\right)y^{\prime}=1,\qquad\text{i.e.}\qquad y^{\prime}\left(1+\sin y\right)=1,\] so \[y^{\prime}=\frac{1}{1+\sin y}\quad (1+\sin y\neq 0).\]Now substitute in the left-hand side of the given equation: \[\left(y\sin y+\cos y+x\right)y^{\prime}=\frac{y\sin y+\cos y+x}{1+\sin y}.\]The step that makes it work is replacing \(\displaystyle x\) by \(\displaystyle y-\cos y\) from the given relation: \[y\sin y+\cos y+x=y\sin y+\cos y+\left(y-\cos y\right)=y\sin y+y=y\left(1+\sin y\right).\]Therefore \[\left(y\sin y+\cos y+x\right)y^{\prime}=\frac{y\left(1+\sin y\right)}{1+\sin y}=y,\] which is the right-hand side.Hence \(\displaystyle y-\cos y=x\) is a solution of \(\displaystyle \left(y\sin y+\cos y+x\right)y^{\prime}=y\).

Differentiate the implicit relation with respect to \(\displaystyle x\) and clear the fraction.Given \(\displaystyle x+y=\tan^{-1}y\). Differentiating both sides with respect to \(\displaystyle x\), using \(\displaystyle \dfrac{d}{dx}\left(\tan^{-1}y\right)=\dfrac{1}{1+y^{2}}\,y^{\prime}\): \[1+y^{\prime}=\frac{y^{\prime}}{1+y^{2}}.\]Multiply both sides by \(\displaystyle 1+y^{2}\) (never zero): \[\left(1+y^{2}\right)\left(1+y^{\prime}\right)=y^{\prime},\] that is, \[1+y^{2}+y^{\prime}+y^{2}y^{\prime}=y^{\prime}.\]Cancelling \(\displaystyle y^{\prime}\) from both sides: \[y^{2}y^{\prime}+y^{2}+1=0.\]This is exactly the given differential equation, so \(\displaystyle x+y=\tan^{-1}y\) is a solution of \(\displaystyle y^{2}y^{\prime}+y^{2}+1=0\).

Differentiate by the chain rule and express the derivative in terms of \(\displaystyle y\).Given \(\displaystyle y=\sqrt{a^{2}-x^{2}}=\left(a^{2}-x^{2}\right)^{1/2}\) for \(\displaystyle x\in(-a,a)\), so that \(\displaystyle a^{2}-x^{2}>0\) and the root is real and positive. By the chain rule, \[\frac{dy}{dx}=\tfrac{1}{2}\left(a^{2}-x^{2}\right)^{-1/2}\cdot\frac{d}{dx}\left(a^{2}-x^{2}\right)=\tfrac{1}{2}\left(a^{2}-x^{2}\right)^{-1/2}\left(-2x\right)=\frac{-x}{\sqrt{a^{2}-x^{2}}}.\] The minus sign coming from \(\displaystyle \frac{d}{dx}\left(-x^{2}\right)\) is the step to be careful with.Since \(\displaystyle y=\sqrt{a^{2}-x^{2}}\) and \(\displaystyle y\neq 0\) on \(\displaystyle (-a,a)\), \[\frac{dy}{dx}=-\frac{x}{y}.\]Substituting in the left-hand side of the equation: \[x+y\frac{dy}{dx}=x+y\left(-\frac{x}{y}\right)=x-x=0,\] which is the right-hand side.Hence \(\displaystyle y=\sqrt{a^{2}-x^{2}},\ x\in(-a,a)\), is a solution of \(\displaystyle x+y\dfrac{dy}{dx}=0\), \(\displaystyle y\neq 0\).

Use the definition: the general solution of a differential equation is the solution containing as many arbitrary constants as the order of the equation.The order of a differential equation is the order of the highest derivative occurring in it. Recovering a function from its \(\displaystyle n\)th derivative requires \(\displaystyle n\) successive integrations, and each integration introduces one arbitrary constant; hence a differential equation of order \(\displaystyle n\) has a general solution with exactly \(\displaystyle n\) arbitrary constants.Here the order is \(\displaystyle 4\), so the general solution contains \(\displaystyle 4\) arbitrary constants.The correct option is \(\displaystyle (\mathrm{D})\ 4\).

Distinguish the general solution from a particular solution.The general solution of a differential equation of order \(\displaystyle n\) contains \(\displaystyle n\) arbitrary constants. A particular solution is obtained from the general solution by assigning particular values to those arbitrary constants (for instance, from given initial or boundary conditions); once the values are fixed, no arbitrary constant remains.So for a third order equation the general solution has \(\displaystyle 3\) arbitrary constants, but a particular solution has none, whatever the order.The correct option is \(\displaystyle (\mathrm{D})\ 0\).