Easy Practice — Set 1

Quick Recap — Order, Degree & Basic Solutions

• Order = order of the highest derivative present; degree = power of that highest derivative (only when the equation is polynomial in derivatives).
• The general solution of an $n$th-order DE contains $n$ arbitrary constants; a particular solution has none.
• Simple integrations: $\tfrac{dy}{dx}=f(x)\Rightarrow y=\int f(x)\,dx+C$.

Beyond-NCERT JEE Formulae

These four devices handle the JEE first-order ODEs that fall outside the NCERT toolkit (separable, homogeneous, and linear-with-$\mathrm{IF}$).

1. Bernoulli's Equation

$$\dfrac{dy}{dx}+P(x)\,y=Q(x)\,y^{n},\qquad n\neq 0,1.$$ Divide throughout by $y^{n}$ and substitute $v=y^{1-n}$, so that $\dfrac{dv}{dx}=(1-n)\,y^{-n}\dfrac{dy}{dx}$. The equation linearises to $$\dfrac{dv}{dx}+(1-n)\,P(x)\,v=(1-n)\,Q(x),$$ now solved by the ordinary integrating factor $\mathrm{IF}=e^{\int (1-n)P\,dx}$. [JEE Tip] Spot Bernoulli the moment a bare power $y^{n}$ sits on the right; do not waste time attempting separation first.

2. Linear in $x$ (swap the roles of the variables)

An equation that is messy in $y$ is often linear when $x$ is treated as the dependent variable: $$\dfrac{dx}{dy}+P(y)\,x=Q(y),\qquad \mathrm{IF}=e^{\int P\,dy},$$ after which $x\cdot\mathrm{IF}=\displaystyle\int Q\cdot\mathrm{IF}\,dy+C$. Use this whenever $y$ enters awkwardly (through $\tan^{-1}y$, $e^{y}$, $\ln y$, …) while $x$ appears only to the first power.

3. Exact-Differential Recognitions (regrouping)

Memorise these total differentials and read them backwards: $$d(xy)=x\,dy+y\,dx,\qquad d\!\left(\dfrac{y}{x}\right)=\dfrac{x\,dy-y\,dx}{x^{2}},\qquad d\!\left(\dfrac{x}{y}\right)=\dfrac{y\,dx-x\,dy}{y^{2}},$$ $$d\!\left(x^{2}+y^{2}\right)=2\,(x\,dx+y\,dy),\qquad d\!\left(\tan^{-1}\dfrac{y}{x}\right)=\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}.$$ [JEE Tip] When you spot the block $x\,dy-y\,dx$ or $x\,dx+y\,dy$, divide by the matching factor ($x^{2}$, $y^{2}$, $x^{2}+y^{2}$, or $2\sqrt{x^{2}+y^{2}}$) to fold the whole equation into a single total differential, then integrate on sight.

4. Substitution $\dfrac{dy}{dx}=f(ax+by+c)$

Put $t=ax+by+c$; then $\dfrac{dt}{dx}=a+b\,\dfrac{dy}{dx}=a+b\,f(t)$, which is always variables-separable: $$\dfrac{dt}{a+b\,f(t)}=dx.$$ Integrate and back-substitute $t=ax+by+c$ at the end.

Solved Examples — Beyond-NCERT Formulae

Example 1 (Bernoulli). Solve $\dfrac{dy}{dx}+\dfrac{y}{x}=y^{2}$.

Here $n=2$. Dividing by $y^{2}$ gives $y^{-2}\dfrac{dy}{dx}+\dfrac{1}{x}\,y^{-1}=1$. Put $v=y^{-1}$, so that $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$, and the equation becomes $$-\dfrac{dv}{dx}+\dfrac{v}{x}=1\quad\Longrightarrow\quad \dfrac{dv}{dx}-\dfrac{v}{x}=-1.$$ With $\mathrm{IF}=e^{-\int dx/x}=\dfrac{1}{x}$, this is $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=-\dfrac{1}{x}$, so $\dfrac{v}{x}=-\ln x+C$. Hence $v=x\,(C-\ln x)$, and since $v=\dfrac{1}{y}$, $$\dfrac{1}{y}=x\,(C-\ln x).$$

Example 2 (Bernoulli). Solve $x\dfrac{dy}{dx}+y=y^{2}\ln x$.

Divide by $x$ to reach standard form $\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{\ln x}{x}\,y^{2}$ ($n=2$). Dividing by $y^{2}$ and putting $v=y^{-1}$ (so $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$), $$\dfrac{dv}{dx}-\dfrac{v}{x}=-\dfrac{\ln x}{x}.$$ Here $\mathrm{IF}=\dfrac{1}{x}$, so $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=-\dfrac{\ln x}{x^{2}}$. Using $\displaystyle\int\dfrac{\ln x}{x^{2}}\,dx=-\dfrac{\ln x}{x}-\dfrac{1}{x}$, we get $\dfrac{v}{x}=\dfrac{\ln x}{x}+\dfrac{1}{x}+C$, i.e. $v=1+\ln x+Cx$. Therefore $$\dfrac{1}{y}=1+\ln x+Cx.$$

Example 3 (Linear in $x$). Solve $(1+y^{2})\,dx=(\tan^{-1}y-x)\,dy$.

The presence of $\tan^{-1}y$ blocks a linear-in-$y$ reading, so divide by $(1+y^{2})\,dy$ and collect $x$: $$\dfrac{dx}{dy}+\dfrac{x}{1+y^{2}}=\dfrac{\tan^{-1}y}{1+y^{2}}.$$ Here $P(y)=\dfrac{1}{1+y^{2}}$, so $\mathrm{IF}=e^{\int dy/(1+y^{2})}=e^{\tan^{-1}y}$. Putting $s=\tan^{-1}y$, the right side integrates by parts as $\displaystyle\int s\,e^{s}\,ds=(s-1)e^{s}$, giving $x\,e^{\tan^{-1}y}=(\tan^{-1}y-1)\,e^{\tan^{-1}y}+C$. Thus $$x=\tan^{-1}y-1+C\,e^{-\tan^{-1}y}.$$

Example 4 (Linear in $x$). Solve $(2x-10y^{3})\dfrac{dy}{dx}+y=0$, with $y>0$.

This is quadratic-looking in $y$ but linear in $x$. Invert the derivative: $$\dfrac{dx}{dy}=\dfrac{2x-10y^{3}}{-y}=-\dfrac{2x}{y}+10y^{2}\quad\Longrightarrow\quad \dfrac{dx}{dy}+\dfrac{2}{y}\,x=10y^{2}.$$ Then $\mathrm{IF}=e^{\int (2/y)\,dy}=y^{2}$, so $\dfrac{d}{dy}\!\left(x\,y^{2}\right)=10y^{4}$, giving $x\,y^{2}=2y^{5}+C$. Hence $$x=2y^{3}+\dfrac{C}{y^{2}}.$$

Example 5 (Regrouping). Solve $x\,dy-y\,dx=(x^{2}+y^{2})\,dx$.

Divide by $x^{2}+y^{2}$ and read the left side as a total differential using $d\!\left(\tan^{-1}\dfrac{y}{x}\right)=\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}$: $$\dfrac{x\,dy-y\,dx}{x^{2}+y^{2}}=dx\quad\Longrightarrow\quad d\!\left(\tan^{-1}\dfrac{y}{x}\right)=dx.$$ Integrating gives $\tan^{-1}\dfrac{y}{x}=x+C$, that is $$y=x\tan(x+C).$$

Example 6 (Regrouping). Solve $x\,dy+y\,dx=xy\,dx$.

The left side is exactly $d(xy)$. Put $u=xy$; the equation becomes $du=u\,dx$, so $\dfrac{du}{u}=dx$ and $\ln|u|=x+C_{1}$. Exponentiating, $u=A\,e^{x}$, i.e. $$xy=A\,e^{x}.$$

Example 7 (Substitution $t=x+y$). Solve $\dfrac{dy}{dx}=(x+y)^{2}$.

Put $t=x+y$, so $\dfrac{dt}{dx}=1+\dfrac{dy}{dx}=1+t^{2}$. This separates as $\dfrac{dt}{1+t^{2}}=dx$, giving $\tan^{-1}t=x+C$. Back-substituting $t=x+y$, $$x+y=\tan(x+C)\quad\Longrightarrow\quad y=\tan(x+C)-x.$$ (For instance $y(0)=0$ forces $C=0$, so $y=\tan x-x$.)

Example 8 (Substitution $t=x+y$). Solve $\dfrac{dy}{dx}=\cos(x+y)$.

Put $t=x+y$, so $\dfrac{dt}{dx}=1+\cos t$. Using $1+\cos t=2\cos^{2}\dfrac{t}{2}$, $$\dfrac{dt}{2\cos^{2}(t/2)}=dx\quad\Longrightarrow\quad \tfrac{1}{2}\int \sec^{2}\dfrac{t}{2}\,dt=\int dx\quad\Longrightarrow\quad \tan\dfrac{t}{2}=x+C.$$ Therefore $$\tan!\left(\dfrac{x+y}{2}\right)=x+C.