Question
If $\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},$ prove that $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0.$
✓
Answer
We have,
$\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},...(1)$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{nx}^{\text{n}-1}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}+\text{x}^\text{n}\\\Big\{-\text{a}\sin(\log\text{x})\times\frac{1}{\text{x}}+\text{b}\cos(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text {x})\}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}$
$\Rightarrow\frac{\text{n}}{\text{x}}\text{y}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\ [\text{from}(1)]$
$\Rightarrow\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}=\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}...(2)$
Differentiating $\frac{\text{dy}}{\text{dx}}$ with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\text{x}^{\text{n}-2}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\+\text{x}^{\text{n}-1}\Big\{-\text{a}\cos(\log\text{x})\times\frac{1}{\text{x}}-\text{b}\sin(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\frac{\text{dx}}{\text{dy}}-\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\frac{\text{x}^{\text{n}-1}}{\text{x}}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\-\frac{\text{x}^\text{n}}{\text{x}^2}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}$
$\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{ny}}{\text{x}^2}+\Big(\frac{\text{n}-1}{\text{x}}\Big)\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}\Big)-\frac{\text{y}}{\text{x}^2}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{n}+\text{n}-1}{\text{x}}\Big)-\frac{(\text{n}+\text{n}^2+\text{n}+1)\text{y}}{\text{x}^2}$
$=\Big(\frac{2\text{n}-1}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{(\text{n}^2+1)\text{y}}{\text{x}^2}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}(2\text{n}-1)\frac{\text{dy}}{\text{dx}}+(\text{n}^2+1)\text{y}=0$
Hence, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\text{x}\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0$
$\text{y}=\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\},...(1)$
Differentiating y with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=\text{nx}^{\text{n}-1}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}+\text{x}^\text{n}\\\Big\{-\text{a}\sin(\log\text{x})\times\frac{1}{\text{x}}+\text{b}\cos(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\text{x}^\text{n}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text {x})\}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}$
$\Rightarrow\frac{\text{n}}{\text{x}}\text{y}+\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\ [\text{from}(1)]$
$\Rightarrow\text{x}^{\text{n}-1}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}=\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}...(2)$
Differentiating $\frac{\text{dy}}{\text{dx}}$ with respect to x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}=\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\text{x}^{\text{n}-2}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\+\text{x}^{\text{n}-1}\Big\{-\text{a}\cos(\log\text{x})\times\frac{1}{\text{x}}-\text{b}\sin(\log\text{x})\times\frac{1}{\text{x}}\Big\}$
$=\frac{\text{n}}{\text{x}}\frac{\text{dx}}{\text{dy}}-\frac{\text{ny}}{\text{x}^2}+(\text{n}-1)\frac{\text{x}^{\text{n}-1}}{\text{x}}\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log\text{x})\}\\-\frac{\text{x}^\text{n}}{\text{x}^2}\{\text{a}\cos(\log\text{x})+\text{b}\sin(\log\text{x})\}$
$\frac{\text{n}}{\text{x}}\frac{\text{dy}}{\text{dx}}-\frac{\text{ny}}{\text{x}^2}+\Big(\frac{\text{n}-1}{\text{x}}\Big)\Big(\frac{\text{dy}}{\text{dx}}-\frac{\text{n}}{\text{x}}\text{y}\Big)-\frac{\text{y}}{\text{x}^2}$
$=\frac{\text{dy}}{\text{dx}}\Big(\frac{\text{n}+\text{n}-1}{\text{x}}\Big)-\frac{(\text{n}+\text{n}^2+\text{n}+1)\text{y}}{\text{x}^2}$
$=\Big(\frac{2\text{n}-1}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\frac{(\text{n}^2+1)\text{y}}{\text{x}^2}$
$\Rightarrow\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}(2\text{n}-1)\frac{\text{dy}}{\text{dx}}+(\text{n}^2+1)\text{y}=0$
Hence, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}+(1-2\text{n})\text{x}\frac{\text{dy}}{\text{dx}}+(1+\text{n}^2)\text{y}=0$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Prove that the curves $y^2=4 x$ and $x^2=4 y$ divide the area of the square bounded by sides $x=0, x=4, y=4$ and $y =0$ into three equal parts.
→Show that the three lines with direction cosines $\frac{12}{13},\frac{-3}{13},\frac{-4}{13},\frac{4}{13},\frac{12}{13},\frac{3}{13},\frac{3}{13},\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular.
→Differentiate $\sin(x^{2} + 1)$with respect to $ x$from first principle.
→Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\cos\text{x}$
→$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\cos\text{x}$
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by $f(\text{x})=\Big(\frac{\text{x}-2}{\text{x}-3}\Big).$ Is f one-one and onto? Justify your answer.
→Using elementary row transformations, find the inverse of the matrix $\text{A}=\begin{bmatrix}1 & 2&3 \\2 & 5&7\\-2&-4&-5 \end{bmatrix}.$
→A manufacturer produces three types of bolts, $x, y$ and $z$ which he sells in two markets. Annual sales (in ₹) are indicated below:
If unit sales prices of $x, y$ and z are ₹ 2.50, ₹ $1.50$ and ₹ $1.00$ respectively, then answer the following questions using the concept of matrices.
→
| Markets | Products | ||
| $X$ | $Y$ | $Z$ | |
| I | $10000$ | $2000$ | $18000$ |
| II | $6000$ | $20000$ | $8000$ |
- Find the total revenue collected from the Market-I.
- $₹\ 44000$
- $₹\ 48000$
- $₹\ 46000$
- $₹\ 53000$
- Find the total revenue collected from the Market-II.
- $₹\ 51000$
- $₹\ 53000$
- $₹\ 46000$
- $₹\ 49000$
- If the unit costs of the above three commodities are $₹\ 2.00, ₹\ 1.00$ and $50$ paise respectively, then find the gross profit from both the markets.
- $₹\ 53000$
- $₹\ 46000$
- $₹\ 34000$
- $₹\ 32000$
- If matrix $\text{A}=[\text{a}_\text{ij}]_{2\times2},$ where $\text{a}_\text{ij}=1,$ if $\text{i}\neq\text{j}$ and $\text{a}_\text{ij}=0,$ if $\text{i}=\text{j}$ then $A^2$ is equal to:
- $I$
- $A$
- $OR$
- None of these
- If $A$ and $B$ are matrices of same order, then $(AB' - BA')$ is a.
- Skew-synunetric matrix.
- Null matrix.
- Symmetric matrix.
- Unit matrix.
Prove that |A adj $A| = |A|^n.$
→A family has 2 children. Find the probability that both are boys, if it is known that.
- at least one of the children is a boy,
- the elder child is a boy.
Let $\vec{\text{a}}=\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}.$Find a vector $\vec{\text{d}}$ which is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{d}}$ and $\vec{\text{c}}.\vec{\text{d}}=15.$
→