Differential Equations - Formation of a Differential Equation and Problem Solving (Lecture-02)
Solved Examples
1. The solution of $x^{3} \dfrac{d y}{d x}+4 x^{2} \tan y=e^{x}$ secy satisfying $y$ ( 1$)=0$, is
(a) $\tan y=(x-2) \mathrm{e}^{x} \cdot \log _{\mathrm{e}} \mathrm{x}$
(b) $\quad \sin y=e^{x}(x-1) x^{-4}$
(c) $\tan y=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}} \mathrm{x}^{-3}$
(d) $\quad \sin y=e^{x}(x-1) x^{-3}$
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Solution
The equation can be written as
$\cos y \dfrac{d y}{d x}+\dfrac{4}{x} \sin y=\dfrac{e^{x}}{x^{3}}$
put $\sin y=z \Rightarrow \operatorname{cosy} \cdot \dfrac{d y}{d x}=\dfrac{d z}{d x}$
$\therefore \dfrac{\mathrm{dz}}{\mathrm{dx}}+\dfrac{4 \mathrm{z}}{\mathrm{x}}=\dfrac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^{3}}$
$\therefore$ I.F $=\mathrm{e}^{\int \dfrac{4}{\mathrm{x}} \mathrm{dx}} 4=\mathrm{x}^{4}$
$\therefore z \cdot x^{4}=\int \mathrm{x}^{4} \cdot \dfrac{\mathrm{e}^{\mathrm{x}}}{\mathrm{x}^{3}} \cdot \mathrm{dx}+\mathrm{C}$
$\Rightarrow \mathrm{zx}^{4}=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{C}$
$\therefore \mathrm{x}^{4} \sin \mathrm{y}=(\mathrm{x}-1) \mathrm{e}^{\mathrm{x}}+\mathrm{C}$ $\hspace {3 cm}$ (put $\mathrm{x}=1, \mathrm{y}=0$)
$\therefore \sin \mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1) \mathrm{x}^{-4}$
Answer (b)
2. The solution of the equation
$\dfrac{d y}{d x}+x(x+y)=x^{3}(x+y)^{3}-1$ is
(a) $\dfrac{1}{x+y}=x^{2}+1+\mathrm{Ce}^{x}$
(b) $\dfrac{1}{(x+y)^{2}}=x^{2}+1+\mathrm{Ce}^{\mathrm{x}^{2}}$
(c) $\dfrac{1}{(x+y)^{2}}=x+1+C e^{x}$
(d) $\dfrac{1}{x+y}=x+1+\mathrm{Ce}^{x^{2}}$
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Solution
$\begin{aligned} & \text { Put } z=\dfrac{1}{(x+y)^{2}} \\ & \text { Now } \dfrac{d z}{d x}=\dfrac{-2}{(x+y)^{3}}\left(1+\dfrac{d y}{d x}\right) \end{aligned}$
$\begin{aligned} & \therefore \dfrac{1}{(\mathrm{x}+\mathrm{y})^{3}}\left(\dfrac{\mathrm{dy}}{\mathrm{dx}}+1\right)=\dfrac{-\mathrm{x}}{(\mathrm{x}+\mathrm{y})^{2}}+\mathrm{x}^{3} \\ & \Rightarrow \dfrac{-1}{2} \dfrac{\mathrm{dz}}{\mathrm{dx}}=-\mathrm{zx}+\mathrm{x}^{3} \\ & \Rightarrow \dfrac{\mathrm{dz}}{\mathrm{dx}}-2 \mathrm{zx}=-2 \mathrm{x}^{3} \\ & \therefore \mathrm{I} \cdot \mathrm{F}=\mathrm{e}^{-\int 2 \mathrm{xdx}}=\mathrm{e}^{-\mathrm{x}^{2}} \\ & \therefore \mathrm{z} \cdot \mathrm{e}^{-\mathrm{x}^{2}}=\int \mathrm{e}^{-\mathrm{x}^{2}} \cdot-2 \mathrm{x}^{3} \cdot \mathrm{dx}+\mathrm{C}=-\int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{tdt}+\mathrm{C} \\ & \Rightarrow \dfrac{1}{(\mathrm{x}+\mathrm{y})^{2}} \mathrm{e}^{-\mathrm{x}^{2}}=-\mathrm{te}^{\mathrm{t}}+\mathrm{e}^{\mathrm{t}}+\mathrm{C} \\ & \Rightarrow \dfrac{1}{(\mathrm{x}+\mathrm{y})^{2}}=\mathrm{x}^{2}+1+\mathrm{e}^{\mathrm{x}^{2}} \end{aligned}$
Answer (b)
3. The solution of differential equation
$x=1+x y \dfrac{d y}{d x}+\dfrac{x^{2} y^{2}}{2 !}\left(\dfrac{d y}{d x}\right)^{2}+\dfrac{x^{3} y^{3}}{3 !}\left(\dfrac{d y}{d x}\right)^{3}+$ is
(a) $y=\log _{e} x+C$
(b) $y=\left(\log _{\mathrm{e}} \mathrm{x}\right)^{2}+\mathrm{C}$
(c) $\mathrm{y}= \pm \log _{\mathrm{e}} \mathrm{x}+\mathrm{C}$
(d) $x y=x^{y}+C$
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Solution
$\begin{aligned} & x=e^{x y \dfrac{d y}{d x}} \Rightarrow \log _{e} x=x y \dfrac{d y}{d x} \\ & \therefore \int y d y=\int \dfrac{1}{x} \log _{e} x d x+C \\ & \dfrac{y^{2}}{2}=\dfrac{\left(\log _{e} x\right)^{2}}{2}+C \\ & \Rightarrow y= \pm \sqrt{\left(\log _{e} x\right)^{2}}+C \\ & \Rightarrow y= \pm \log _{e} x+C \end{aligned}$
Answer (c)
4. The function $\mathrm{f}(\mathrm{x})$ satisfying $(\mathrm{f}(\mathrm{x}))^{2}+4 \mathrm{f}(\mathrm{x}) \cdot \mathrm{f}^{\prime}(\mathrm{x})+\left(\mathrm{f}^{\prime}(\mathrm{x})\right)^{2}=0$ is given by
(a) $\mathrm{ke}^{(2+\sqrt{3}) \mathrm{x}}$
(b) $\mathrm{ke}^{(4 \pm \sqrt{5}) \mathrm{x}}$
(c) $\mathrm{ke}^{(-2 \pm \sqrt{3}) \mathrm{x}}$
(d) None of these
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Solution
Solving we get
$f^{\prime}(x)=\dfrac{-4 f(x) \pm \sqrt{16(f(x))^{2}-4(f(x))^{2}}}{2.1}=-2 f(x) \pm \sqrt{3} f(x)$
$\begin{aligned} & \Rightarrow \dfrac{\mathrm{f}^{1}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}=-2 \pm \sqrt{3} \\ & \log _{\mathrm{e}} \mathrm{f}(\mathrm{x})=(-2 \pm \sqrt{3}) \mathrm{x}+\mathrm{C} \\ & \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{e}^{(-2 \pm \sqrt{3}) \mathrm{x}+\mathrm{C}}=\mathrm{e}^{(-2 \pm \sqrt{3}) \mathrm{x}} \quad \mathrm{e}^{\mathrm{C}}=\mathrm{ke}^{(-2 \pm \sqrt{3}) \mathrm{x}} \text { where } \mathrm{k}=\mathrm{e}^{\mathrm{C}} \end{aligned}$
Answer (c)
5. A curve $f(x)$ has normal at the point $P(1,1)$ given by $a(y-1)+(x-1)=0$. If the slope of tangent at any point on the curve is proportional to the ordinate at that point, then the curve is
(a) $y=e^{a x}-1$
(b) $\mathrm{y}-1=\mathrm{e}^{\mathrm{ax}}$
(c) $y=e^{a(x-1)}$
(d) None of these
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Solution
Normal at point $\mathrm{P}$ is ay $+\mathrm{x}=\mathrm{a}+1$
$\therefore$ Slope of tangent at $\mathrm{P}=\mathrm{a}=\left(\dfrac{\mathrm{dy}}{\mathrm{dx}}\right)(1,1)$……………………..(1)
Now $\dfrac{d y}{d x} \alpha y \Rightarrow \dfrac{d y}{d x}=k y \Rightarrow\left(\dfrac{d y}{d x}\right) _{(1,1)}=k$………………….(2)
From (1) & (2) $\mathrm{k}=\mathrm{a}$
$\therefore \dfrac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ay} \Rightarrow \dfrac{\mathrm{dy}}{\mathrm{y}}=\mathrm{a} \cdot \mathrm{dx}$
$\Rightarrow \log _{\mathrm{e}} \mathrm{y}=\mathrm{ax}+\mathrm{C}$
Now $\mathrm{C}=-\mathrm{a}($ as curve passes through $(1,1))$
$\therefore \log _{\mathrm{e}} \mathrm{y}=\mathrm{ax}-\mathrm{a}$
$\Rightarrow \mathrm{y}=\mathrm{e}^{\mathrm{a}(\mathrm{x}-1)}$
Answer (c)
6. If the length of portion of normal intercepted between the curve \& $x$-axis varies as the square of the ordinate, then the curve is
(a) $\mathrm{kx}+\sqrt{1-\mathrm{k}^{2} \mathrm{x}^{2}}=$ C. $\mathrm{e}^{\mathrm{ky}}$
(b) $\mathrm{ky}+\sqrt{\mathrm{k}^{2} \mathrm{y}^{2}-1}=\mathrm{C} \cdot \mathrm{e}^{\mathrm{kx}}$
(c) $\mathrm{ky}+\sqrt{\mathrm{k}^{2} \mathrm{x}^{2}-1}=\mathrm{Ce}^{\mathrm{xy}}$
(n) None of these
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Solution
Normal is $Y-y=-\dfrac{d x}{d y}(X-x)$
Point on $\mathrm{x}$-axis is $\left(\mathrm{x}+\mathrm{y} \dfrac{\mathrm{dy}}{\mathrm{dx}}, 0\right)$
Now $\sqrt{\left(y \dfrac{d y}{d x}+x-x\right)^{2}+(0-y)^{2}} \alpha y^{2}$
$\Rightarrow \sqrt{\left(y \dfrac{d y}{d x}\right)^{2}+y^{2}}=k y^{2}$
$\Rightarrow \dfrac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{\mathrm{k}^{2} \mathrm{y}^{2}-1}$
$\Rightarrow \dfrac{\mathrm{dy}}{\sqrt{\mathrm{k}^{2} \mathrm{y}^{2}-1}}=\mathrm{dx}$
Integrate to get,
$\begin{aligned} & \text { loge }\left|\mathrm{y}+\sqrt{\mathrm{y}^{2}-\dfrac{1}{\mathrm{k}^{2}}}\right|=\mathrm{kx}+\mathrm{C} \\ & \Rightarrow \mathrm{ky}+\sqrt{\mathrm{k}^{2} \mathrm{y}^{2}-1}=\mathrm{C} \cdot \mathrm{e}^{\mathrm{kx}} \end{aligned}$
Answer (b)
Exercise
1. Let $y^{1}(x)+y(x) g^{1}(x)=g(x) g^{1}(x), y(0)=0, x \in R$, where $f^{1}(x)$ denotes $\dfrac{d}{d x} f(x)$ and $g(x)$ is a given non-constant differentiable function on $R$ with $g(0)=g(2)=0$. Then the value of $y(2)$ is
(a) 2
(b) -1
(c) 0
(d) None of these
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Answer: c2. The solution of the differential equation $\dfrac{d^{3} y}{d x^{3}}-8 \dfrac{d^{2} y}{d x^{2}}=0$ satisfying $y(0)=\dfrac{1}{8}, y _{1}(0)=0$ and $y _{2}(0)=1$ is $y _{1}=\dfrac{e^{8 x}-8 x+7}{\lambda}$, then the numerical value of $\lambda$ is
(a) 8
(b) $\dfrac{7}{64}$
(c) 64
(d) $\dfrac{-1}{8}$
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Answer: c3.* If the solution of the differential equation $\dfrac{d y}{d x}+\dfrac{\cos x(3 \cos y-7 \sin x-3)}{\sin y(3 \sin x-7 \cos y+7)}=0$ is $(\sin x+\cos y-1)^{\lambda} .(\sin x-\cos y+1)^{\mu}=c$, where $c$ is arbitrary constant, then
(a) $\lambda=5$
(b) $\lambda=2$
(c) $\mu=2$
(d) $\mu=5$
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Answer: a,c4.* The curve for which the area of the triangle formed by the $\mathrm{x}$-axis, the tangent line and radius vector of the point of tangency is equal to $\mathrm{a}^{2}$ is
(a) $\mathrm{x}=\mathrm{Cy}+\dfrac{\mathrm{a}^{2}}{\mathrm{y}}$
(b) $y=x-C x^{2}$
(c) $\mathrm{y}=\mathrm{Cx}+\dfrac{\mathrm{a}^{2}}{\mathrm{x}}$
(d) $x=C y-\dfrac{a^{2}}{y}$
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Answer: a,d5.* The solution of $\left(\dfrac{x d x+y d y}{x d y-y d x}\right)=\sqrt{\dfrac{a^{2}-x^{2}-y^{2}}{x^{2}+y^{2}}}$ is
(a) $\sqrt{x^{2}+y^{2}}=\operatorname{asin}\left(\left(\tan ^{-1} y / x\right)+\right.$ constant $)$
(b) $\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}=\operatorname{acos}\left(\left(\tan ^{-1} \mathrm{y} / \mathrm{x}\right)+\right.$ constant $)$
(c) $\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}=\mathrm{a}\left(\tan \left(\sin ^{-1} \mathrm{y} / \mathrm{x}\right)+\right.$ constant $)$
(d) $y=x \tan \left(\right.$ constant $\left.+\sin ^{-1} \dfrac{1}{a} \sqrt{x^{2}+y^{2}}\right)$
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Answer: a,d6. If $f(x) \int(f(x))^{-2} d x$ and $f(1)=0$, then $f(x)$ is equal to
(a) $(2(x-1))^{1 / 4}$
(b) $(5(x-2))^{1 / 5}$
(c) $(3(x-1))^{1 / 3}$
(d) None of these
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Answer: c7. Through any point ( $\mathrm{x}, \mathrm{y})$ of a curve which passes through the origin, lines are drawn parallel to the coordinate axes. The curve, given that it divedes the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of
(a) circle
(b) parabola
(c) ellipse
(d) hyperbola
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Answer: b8. Match the following
| Column I | Column II | ||
|---|---|---|---|
| (a) | If the function $y=e^{4 x}+2 e^{-x}$ is a solution of the differential | (p) | 3 |
| equation $\dfrac{d^{3} y}{d x^{3}}-13 \dfrac{d y}{d x}=k y$, then the value of $\dfrac{k}{3}$ is | |||
| (b) | Number of straight lines which satisfy the differential | (q) | 4 |
| equation $\dfrac{d y}{d x}+x\left(\dfrac{d y}{d x}\right)^{2}-y=0$ is | |||
| (c) | If real value of $\mathrm{m}$ for which the substitution, $\mathrm{y}=\mathrm{u}^{\mathrm{m}}$ will | (r) | 2 |
| transform the differential equation, $2 x^{4} y \dfrac{d y}{d x}+y^{4}=4 x^{6}$ | |||
| into a homogenious equation, then the value of $2 \mathrm{~m}$ is | |||
| (d) | If the solution of differential equation $\dfrac{x^{2} d^{2} y}{d x^{2}}+2 x \dfrac{d y}{d x}=12 y$ is | (s) | 1 |
| $\mathrm{y}=\mathrm{Ax}^{\mathrm{m}}+\mathrm{Bx}^{-n}$, then $|\mathrm{m}+\mathrm{n}|$ is |
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Answer: $\mathrm{a} \rightarrow \mathrm{q} ; \mathrm{b} \rightarrow \mathrm{r} ; \mathrm{c} \rightarrow \mathrm{p} ; \mathrm{d} \rightarrow \mathrm{s}$9. Read the following passage and answer the questions
If any differential equation be in the form $\mathrm{f}\left(\mathrm{f} _{1}(\mathrm{x}, \mathrm{y}) \mathrm{d}\left(\mathrm{f} _{1}(\mathrm{x}, \mathrm{y})+\phi\left(\mathrm{f} _{2}(\mathrm{x}, \mathrm{y})\right) \mathrm{d}\left(\mathrm{f} _{2}(\mathrm{x}, \mathrm{y})\right)+\ldots \ldots=0\right.\right.$, then we can perform term by term integration. For e.g., $\int \sin x y d(x y)+\int\left(\dfrac{x}{y}\right) d\left(\dfrac{x}{y}\right)=$ $-\operatorname{cosxy}+\dfrac{1}{2}\left(\dfrac{\mathrm{x}}{\mathrm{y}}\right)^{2}+\mathrm{C}$
(i) The solution of the differential equation $\left(x y^{4}+y\right) d x-x d y=0$ is
(a) $\dfrac{\mathrm{x}^{2}}{4}+\dfrac{1}{2}\left(\dfrac{\mathrm{x}}{\mathrm{y}}\right)^{2}=\mathrm{C}$
(b) $\dfrac{\mathrm{x}^{4}}{4}+\dfrac{1}{3}\left(\dfrac{\mathrm{x}}{\mathrm{y}}\right)^{3}=\mathrm{C}$
(c) $\dfrac{\mathrm{x}^{4}}{4}-\dfrac{1}{2}\left(\dfrac{\mathrm{x}}{\mathrm{y}}\right)^{2}=\mathrm{C}$
(d) $\dfrac{x^{2}}{4}-\dfrac{1}{2}\left(\dfrac{x}{y}\right)^{2}=C$
(ii) Solution of differential equation $\left(2 x \cos y+y^{2} \cos x\right) d x+\left(2 y \sin x-x^{2} \sin y\right) d y=0$ is
(a) $x^{2} \cos y+y^{2} \sin x=C$
(b) $x \cos y-y \sin x=C$
(c) $x^{2} \cos ^{2} y+y^{2} \sin ^{2} y=C$
(d) None of these
(iii) Solution of differential equation $\dfrac{x^{2}}{\dfrac{1}{x^{2}}+y^{2}}(x d y+y d x)+y^{2}(x d y-y d x)=0$ is
(a) $\tan ^{-1} x y-\dfrac{1}{2}\left(\dfrac{y}{x}\right)^{2}=C$
(b) $\tan ^{-1} x y+\dfrac{1}{2}\left(\dfrac{y}{x}\right)^{2}=C$
(c) $\tan ^{-1} x y+\dfrac{1}{3}\left(\dfrac{y}{x}\right)^{2}=\mathrm{C}$
(d) None of these
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Answer: (i) b (ii) b (iii) c10. Solution of the differentia equation $\left(e^{x^{2}}+e^{y^{2}}\right)$ y $\dfrac{d y}{d x}+e^{x^{2}}\left(x y^{2}-x\right)=0$ is
(a) $\mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{y}^{2}-1\right)+\mathrm{e}^{\mathrm{y}^{2}}=\mathrm{C}$
(b) $\mathrm{e}^{\mathrm{y}^{2}}\left(\mathrm{x}^{2}-1\right)+\mathrm{e}^{\mathrm{x}^{2}}=\mathrm{C}$
(c) $\mathrm{e}^{\mathrm{y}^{2}}\left(\mathrm{y}^{2}-1\right)+\mathrm{e}^{\mathrm{x}^{2}}=\mathrm{C}$
(d) $\mathrm{e}^{\mathrm{x}^{2}}(\mathrm{y}-1)+\mathrm{e}^{\mathrm{y}^{2}}=\mathrm{C}$
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Answer: a11. The solution of the differential equation $(x \operatorname{coty}+\log \cos x) d y+(\log \sin y-y \tan x) d x=0$
(a) $(\sin x)^{y}(\cos y)^{x}=C$
(b) $(\sin y)^{x}(\cos x)^{y}=C$
(c) $(\sin x)^{x}(\cos y)^{y}=C$
(d) None of these
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Answer: b12. The solution of differential equation
$\dfrac{x+y \dfrac{d y}{d x}}{y-x \dfrac{d y}{d x}}=\dfrac{x \cos ^{2}\left(x^{2}+y^{2}\right)}{y^{3}}$ is
(a) $\tan \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\dfrac{\mathrm{x}^{2}}{\mathrm{y}^{2}}+C$
(b) $\cot \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\dfrac{\mathrm{x}^{2}}{\mathrm{y}^{2}}+\mathrm{C}$
(c) $\tan \left(\mathrm{x}^{3}+\mathrm{y}^{3}\right)=\dfrac{\mathrm{y}^{2}}{\mathrm{x}^{2}}+\mathrm{C}$
(d) $\cot \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\dfrac{\mathrm{y}^{2}}{\mathrm{x}^{2}}+\mathrm{C}$
