Maths M.C.Q (1 Marks)

3. $\{\text{f}(\text{t})+\text{f}(\text{t})\}^2$

Solution:

Here,

$\text{x}=\text{f}(\text{t})\cos\text{t}-\text{f}'(\text{t})\sin\text{t}$

$\text{and}\ \text{y}=\text{f}(\text{t})\sin\text{t}+\text{f}'(\text{t})\cos\text{t}$

$ \Rightarrow\frac{\text{dx}}{\text{dt}}=\text{f}'(\text{t})\cos\text{t}-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}-\text{f}'(\text{t})\cos\text{}\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}'(\text{t})\sin\text)\cos\text{t}+\text{f}'{t}+\text{f}(\text{t})$

$\Rightarrow\frac{\text{dx}}{\text{dt}}=-\text{f}(\text{t})\sin\text{t}-\text{f}''(\text{t})\sin\text{t}$

$ \text{and}\ \frac{\text{dy}}{\text{dt}}=\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}$

Thus

$\Big(\frac{\text{dx}}{\text{dt}}\Big)^2+\Big(\frac{\text{dy}}{\text{dt}}\Big)^2=\{-\text{f}(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\Big\}^2$

$=\{\text{f}(\text{t})\sin\text{t}+\text{f}''(\text{t})\sin\text{t}\}^2+\{\text{f}(\text{t})\cos\text{t}+\text{f}''(\text{t})\cos\text{t}\}^2$

$=\sin^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2+\cos^2\text{t}\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2(\sin^2\text{t}+\cos^2\text{t})$

$=\{\text{f}(\text{t})+\text{f}''(\text{t})\}^2$

2. $\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

Solution:
$\text{y}=\text{a}+\text{bx}^2$

$\Rightarrow\text{y}_1=2\text{bx}$

$\Rightarrow\text{y}_2=2\text{b}$

Multiply by x on both sides we get

$\text{xy}_2=2\text{bx}=\text{y}_1$

$\Rightarrow\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{y}_1$

3. 0

Solution:

$\text{y}=\tan^{-1}\Big\{\frac{\log(\frac{\text{e}}{\text{x}})^2}{\log(\frac{\text{e}}{\text{x}})^2}\Big\}+\tan^{-1}\Big(\frac{3-2\log,\text{x}}{1-6\log,\text{x}}\Big)$

$=\tan^{-1}\Big\{\frac{1-2\log_\text{e}\text{x}}{1+2\log_\text{e}\text{x}}\Big\}+\tan^{-1}\Big\{\frac{3+2\log_\text{e}\text{x}}{1-6\log_\text{e}\text{x}}\Big\}$

$=\tan^{-1}1-\tan^{-1}(2\log_\text{e}\text{x})+\tan^{-1}(3)+\tan^{-1}(2\log_\text{e}\text{x})$

$=\tan^{-1}+\tan^{-1}(3)$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

1. -(n - 1)²y

Solution:

Here,

$\text{y}=\text{x}^{\text{n}-1}\log\text{x}$

$\Rightarrow\text{y}_1=(\text{n}-1)\text{x}^{\text{n-2}}\log\text{x}+\frac{\text{x}^{\text{n}-1}}{\text{x}}$

$\Rightarrow\text{y}_1=\frac{(\text{n}-1)\text{x}^{\text{n}-1}\log\text{x}+\text{x}^{\text{n}-1}}{\text {x}}$

$\Rightarrow\text{xy}_1=(\text{n}-1)\text{y}+\text{x}^{\text{n-1}}$

$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n-2}}$

$\Rightarrow\text{xy}_2+\text{y}_1=(\text{n}-1)\text{y}_1+\frac{(\text{n}-1)\text{x}^{\text{n}-1}}{\text{x}}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{x}^{\text{n}-1}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\{\text{xy}_1-(\text{n}-1)\text{y}\}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)\text{xy}_1-(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=2\text{x}(\text{n}-1)\text{y}_1+(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1(1-2\text{n}+2)=-(\text{n}-1)^2\text{y}$

$\Rightarrow\text{x}^2\text{y}_2+(3-2\text{n})\text{xy}_1=-(\text{n}-1)^2\text{y}$

1. 3 (xy₂ + y₁) y₂

Solution:

$\text{y}=\frac{\text{ax}+\text{b}}{\text{x}^2+\text{c}}$

$\Rightarrow(\text{x}^2+\text{c})\text{y}=\text{ax}+\text{b}$

Differentiating w.r.t.x, we get

$2\text{xy}+(\text{x}^2+\text{c})\frac{\text{dy}}{\text{dx}}=\text{a}$

Differentiating w.r.t.x, we get

$2\text{y}+2\text{xy}_1+2\text{xy}+(\text{x}^2+\text{c})\text{y}_2=0$

$\Rightarrow2\text{y}+4\text{xy}_1+\text{x}^2+\text{cy}_2=0$

Differentiating w.r.t.x, we get

$2\text{y}_1+4\text{y}_1+4\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3+2\text{xy}_2=0$

$\Rightarrow6\text{y}_1+6\text{xy}_2+(\text{x}^2+\text{c})\text{y}_3=0$

$\Rightarrow6\text{y}_1+6\text{xy}_2+\Big(\frac{-2\text{y}-4\text{xy}_1}{\text{y}_2}\Big)\text{y}_3=0$ $[\because2\text{y}+4\text{xy}_1+(\text{x}^2+\text{c})\text{y}_2=0]$

$\Rightarrow6\text{y}_1\text{y}_2+6\text{x}(\text{y}_2)^2-2\text{y}-4\text{xy}_1\text{y}_3=0$

$\Rightarrow3\text{y}_1\text{y}_2+3\text{x}(\text{y}_2)^2-\text{y}-2\text{xy}_1\text{y}_3=0$

$\Rightarrow(\text{y}_1+\text{xy}_2)3\text{y}_2=(2\text{xy}_1+\text{y})\text{y}_3$

2. $2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

Solution:

$\text{y}=2\cos\text{x}\cos3\text{x}=\cos(3\text{x}-\text{x})+\cos(3\text{x}+\text{x})=\cos2\text{x}+\cos4\text{x}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=-2\sin2\text{x}-4\sin4\text{x}=-2(\sin2\text{x}+2\sin4\text{x})$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-4\cos2\text{x}-16\cos4\text{x}=-2^2(\cos2\text{x}+2^2\cos4\text{x})$

$\Rightarrow\frac{\text{d}^3\text{y}}{\text{dx}^3}=2^3(\sin2\text{x}+2^3\sin4\text{x})$

$\Rightarrow\frac{\text{d}^4\text{y}}{\text{dx}^4}=2^3(2\cos2\text{x}+4\times2^3\cos4\text{x})=2^4(\cos2\text{x}+2^4\cos4\text{x})$

$\therefore\frac{\text{d}^{20}(\cos2\text{x}+\cos4\text{x})}{\text{dx}^{20}}=2^{20}(\cos2\text{x}+2^{20}\cos4\text{x})$

1. xy₁ + 2

Solution:

Here,

$\text{y}=(\sin^{-1}\text{x})\frac{1}{\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)^\frac{3}{2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{2\text{x}\sin^{-1}\text{x}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_2=\frac{2}{1-\text{x}^2}+\frac{\text{xy}_1}{(1-\text{x}^2)}$

$\Rightarrow\text{y}_2(1-\text{x}^2)=2+\text{xy}_1$

4. $\frac{3\text{t}}{2}$

Solution:

$\text{x}=\text{t}^2\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}$

$\text{y}=\text{t}^3\Rightarrow\frac{\text{dy}}{\text{dt}}=3\text{t}^2$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}^2}{2\text{t}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}}{2}$

1. $\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$

Solution:

$\text{x}=\text{f}(\text{t})$ $\text{y}=\text{g}(\text{t}),$

$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{g}'}{\text{f}'}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)\frac{\text{dt}}{\text{dx}}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{\text{f}'^2}\frac{1}{\text{f}'}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$​​​​​​​

1. n²y

Solution:

$\text{y}^\frac{1}{\text{n}}+\text{y}-^\frac{1}{\text{n}}=2\text{x}$

Differentiating both sides we get

$\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}-1}-\text{y}^{\frac{1}{\text{n}}-1}\Big)=2$

$\Rightarrow\text{y}_1\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)=2\text{ny}$

Again differentiating both sides we get

$\text{y}_2\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}}\Big)+\frac{\text{y}_1}{\text{n}}\Big(\text{y}^{\frac{1}{\text{n}}}-\text{y}^{\frac{-1}{\text{n}}-1}\Big)=2\text{ny}_1$

$\Rightarrow\text{ny}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+\frac{\text{y}^2_1}{\text{y}}\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)=2\text{n}^2\text{y}_1$

$\Rightarrow\text{nyy}_2\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$

$\Rightarrow\text{nyy}_2\frac{2\text{ny}}{\text{y}_1}+2\text{xy}_1^2=2\text{n}^2\text{yy}_1$

$\Rightarrow\frac{\text{n}^2\text{y}^2\text{y}_2}{\text{y}_1^2}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{\Big(\text{y}^\frac{1}{\text{n}}-\text{y}^\frac{-1}{\text{n}}\Big)^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow\text{y}_2\frac{4\text{x}^2-4}{4}+\text{xy}_1=\text{n}^2\text{y}$

$\Rightarrow(\text{x}^2-1)\text{y}_2+\text{xy}_1=\text{n}^2\text{y}$

3. $\frac{\text{n}!}{(\text{n}-\text{r})!}$

Solution:

$\frac{\text{d}}{\text{dx}}[\text{x}^\text{n}-\text{a}_1\text{x}^{\text{n}-1}+\text{a}_2\text{x}^{\text{n}-2}+...+(-1)^\text{n}\text{a}_\text{n }]\text{e}^\text{x}=\text{x}^\text{n}\ \text{e}^\text{x},$

$\Rightarrow\text{e}^\text{x}(\text{nx}^{\text{n-1}}-\text{a}_1(\text{n}-1)\text{x}^{\text{n-2}}+\text{a}_2(\text{n}-2)\text{x}^{\text{n}-3}+...+(-1)^{\text{n}-1}\text{a}_{\text{n}-1}+\text{x}^\text{a}-\text{a}_1\text{x}^{\text{n-2}}+...+(-1)^\text{n}\text{a}_\text{n})=\text{x}^\text{n}\text{e}^\text{x}$

$\Rightarrow\text{e}^\text{x}(\text{x}^\text{n}+(\text{n}-\text{a}_1)\text{x}^{\text{n}-1}-(\text{a}_1(\text{n-1})-\text{a}_2)\text{x}^{\text{n}-2}\\+(\text{a}_2(\text{n}-2)-\text{a}_3)\text{x}^{\text{n}-3}-...)=\text{x}^\text{n}\text{e}^\text{x}$

on comparing both sides we get

$\text{n}-\text{a}_1=0$

$\Rightarrow\text{a}_1=\text{n}$

$\text{a}_1(\text{n}-1)-\text{a}_2=0$

$\Rightarrow\text{a}_2=\text{a}_1(\text{n}-1)=\text{n}(\text{n}-1)$

$\text{a}_2(\text{n}-2)-\text{a}_3=0$

$\Rightarrow\text{a}_3=\text{a}_2(\text{n}-2)=\text{n}(\text{n}-1)(\text{n}-2)$

So,

$\text{a}_\text{r}=\text{n}(\text{n-1})(\text{n}-2)...(\text{n}-(\text{r}-1)=\frac{\text{n}!}{(\text{n}-\text{r})!}$

1. 1

Solution:

Here,

$\text{f}(\text{x})=\frac{\sin^{-1}\text{x}}{\sqrt{1-\text{x}}^2}$

$\Rightarrow\sqrt{1-\text{x}}^2\text{f}(\text{x})=\sin^{-1}\text{x}$

Differentiating w.r.t.x, we get

$\sqrt{1-\text{x}^2}\text{f}'(\text{x})-\frac{\text{x f}{(\text{x})}}{\sqrt{1-\text{x}}^2}=\frac{1}{\sqrt{1-\text{x}}^2}$

$\Rightarrow(1-\text{x}^2)\text{f}'(\text{x})-\text{xf}(\text{x})=1$

1. ​​​​​​$\frac{1}{2}\text{a}$

Solution:

Here,

$\text{x}=2\text{ at},\text{y}=\text{at}^2,$

Differentiating w.r.t.x, we get

$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$

Differentiating w.r.t.x, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$

Now, $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$

3. $-\Big\{\frac{\text{n}(\text{n}+1)}{2}\Big\}^2$

Solution:

$\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos2\text{x}+\text{i}\sin2\text{x})...(\cos\text{nx}+\text{i}\sin\text{nx})$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})(\cos\text{x}+\text{i}\sin\text{x})^2...(\cos\text{x}+\text{i}\sin\text{x})^\text{n}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{1+2+3........\text{n}}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^{\frac{\text{n}(\text{n}+1)}{2}}$

$\Rightarrow\text{f}(\text{x})=(\cos\text{x}+\text{i}\sin\text{x})^\text{a}$

$\Rightarrow\text{f}(\text{ax})=(\cos\text{ax}+\text{i}\sin\text{ax})...1$

$\Rightarrow\text{f}(1)=(\cos\text{a}+\text{i}\sin\text{a})$

$\Rightarrow1=(\cos\text{a}+\text{i}\sin\text{a})...2\ [\because\text{f}(1)=1]$

Differentiating eqn.1, we get

$\text{f}'(\text{x})=\text{a}(-\sin\text{ax}+\text{i}\cos\text{ax})$

$\Rightarrow\text{f}''(\text{x})=\text-{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''\text{x}=\text{a}^2(-\cos\text{ax}-\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''(\text{x})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{ax}+\text{i}\sin\text{ax})$

$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}(\cos\text{a}+\text{i}\sin\text{a})$

$\Rightarrow\text{f}''({1})=-\Big\{\frac{\text{n}(\text{n}+1)^2}{2}\Big\}\ [\text{using }2]$

2. -n²x

Solution:

Here

$\text{x}=\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$

Differentiating w.r.t.t, we get

$\frac{\text{dx}}{\text{dt}}=-\text{an}\sin\text{nt}-\text{bn}\cos\text{nt}$

Differentiating w.r.t.t, we get

$\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{an}^2\cos\text{nt}+\text{bn}^2\sin\text{nt}$

$=-\text{n}^2\text{a}\cos\text{nt}-\text{b}\sin\text{nt}$

$=-\text{n}^2\text{x}$

3. a function of y only

Solution:

$\text{y}^2=\text{ax}^2+\text{bx}+\text{c}$

$\frac{\text{dy}}{\text{dx}}=2\text{ax}+\text{b}$

$\frac{\text{d}^2\text{y}}{\text{d}^2}=2\text{a}$

$=\text{y}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=2\text{ay}^3$

= A function of y only

3. $(1+\sin2\text{x})\text{y}_1$

Solution:

$\text{y}=\text{e}^{{\tan}\text{x}},$

$\text{y}_1=\text{sec}^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_1=\text{e}^{\tan\text{x}}$

again differentiating w.r.t.x, we get

$\cos^2\text{xy}_2-2\cos\text{x}\sin\text{xy}_1=\sec^2\text{xe}^{\tan\text{x}}$

$\Rightarrow\cos^2\text{xy}_2=\text{y}_1\sin2\text{x}+\text{y}_1$

2. $\text{y}_2=\frac{\text{b}\sin\text{x}}{(\text{a}+\text{b}\cos\text{x})^2}$

4. $-\frac{1}{2\text{at}^3}$

Solution:

$\text{x}=\text{at}^2,\text{y}=2\text{at}$

$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{2\text{a}}{2\text{at}}$

$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\frac{\frac{\text{d}}{\text{dt}}\Big(\frac{\text{dy}}{\text{dx}}\Big)}{\frac{\text{dx}}{\text{dt}}}=\frac{\frac{-1}{\text{t}^2}}{2\text{at}}=\frac{-1}{2\text{at}^3}$

1. $(\text{xy}_1-\text{y})^2$

​​​​​​​Solution:

$\text{y}=\log_\text{e}\Big(\frac{\text{x}}{\text{a}+\text{b}}\Big)^\text{x}$

$\text{y}=\text{x}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))$

$\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\text{x}\Big(\frac{1}{\text{x}}-\frac{\text{b}}{\text{a}+\text{bx}}\Big)$

$=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+1-\frac{\text{bx}}{\text{a}+\text{bx}}$

$(\text{a}+\text{bx})\frac{\text{dy}}{\text{dx}}=(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))(\text{a}+\text{bx})+\text{a}$

Again differentiating w.r.t.x, we get

$(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))\\+(\text{a}+\text{bx})\Big(\frac{1}{\text{x}}\frac{\text{b}}{\text{a}+\text{bx}}\Big)$

$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\text{b}(\log_\text{e}\text{x}-\log_\text{e}(\text{a}+\text{bx}))+\frac{\text{a}}{\text{x}}$

$\Rightarrow(\text{a}+\text{bx})\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{b}\frac{\text{dy}}{\text{dx}}=\frac{\text{by}}{\text{x}}+\frac{\text{a}}{\text{x}}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} =\frac{\text{a}+\text{by}}{\text{x}(\text{a}+\text{bx})}-\frac{\text{b}}{(\text{a}+\text{bx})}\Big(\frac{\text{y}}{\text{x}}+\frac{\text{a}}{\text{a}+\text{bx}}\Big)$

$=\frac{(\text{a}+\text{by})(\text{a}+\text{bx})-\text{b}(\text{ay}+\text{bxy}+\text{ax})}{\text{x}(\text{a}+\text{b})^2}$

$=\frac{\text{a}^2+\text{abx}+\text{aby}+\text{b}^2\text{xy}-\text{bay}-\text{b}^2\text{xy}-\text{abx}}{\text{x}(\text{a}+\text{bx})^2}$

$=\frac{\text{a}^2}{\text{x}(\text{a}+\text{bx})^2}$

$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}$

$(\text{xy}_1-\text{y})=\frac{\text{ax}}{\text{a}+\text{bx}}$

$\text{x}^3\text{y}_2=\frac{\text{x}^2\text{a}^2}{(\text{a}+\text{bx})^2}=(\text{xy}_1-\text{y})^2$

3. -m²y

Solution:

Here,

$\text{y}=\sin(\text{m}\sin^{-1}\text{x}),$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}}{\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{mx}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)^\frac{3}{2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xm}\cos(\text{m}\sin^{-1}\text{x})}{(1-\text{x}^2)\times\sqrt{1-\text{x}^2}}$

$\Rightarrow\text{y}_1=\cos(\text{m}\sin^{-1}\text{x})\frac{\text{m}^2}{(1-\text{x}^2)}+\frac{\text{xy}_1}{(1-\text{x}^2)}$

$\Rightarrow(1-\text{x}^2)\text{y}_2=-\text{ym}^2+\text{xy}_1$

$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1=-\text{m}^2\text{y}$

2. n(n + 1) y

Solution:

Here

$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$

$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$

$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$

$=\text{n}(\text{n}+1)\text{y}$

1. -m²y

Solution:

$\text{y}=\text{a}\sin\text{mx}+\text{b}\cos\text{mx}$

$\frac{\text{dy}}{\text{dx}}=\text{am}\cos\text{mx}-\text{bm}\sin\text{mx}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{am}^2\sin\text{mx}-\text{bm}^2\cos\text{mx}=-\text{m}^2\text{y}$

3. 3

Solution:

$\text{xy}-\log_\text{e}\text{y}=1$

$\Rightarrow\text{xy}_1+\text{y}-\frac{\text{y}_1}{\text{y}}=0$

$\Rightarrow\text{xyy}_1+\text{y}^2-\text{y}_1=0$

$\Rightarrow\text{yy}_1+\text{xy}_1\text{y}_1+\text{xyy}_2+2\text{yy}_1-\text{y}_2=0$

$\Rightarrow\text{x}(\text{y}_1^2+\text{yy}_2)-\text{y}_2+3\text{yy}_2=0$

$\therefore\lambda=3$

4. $\text{f}(\text{e}^\text{x})$

Solution:

Let $\text{y}=\text{f}(\text{e}^\text{x}),$ then

$\text{y}_1=\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

$\text{y}_2=\text{f}''(\text{e}^\text{x})\text{e}^\text{x}\text{e}^\text{x}+\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

$=\text{e}^{2\text{x}}\text{f}''(\text{e}^\text{x})+\text{f}'(\text{e}^\text{x})\text{e}^\text{x}$

3. -y

Solution:

$\text{y}=\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})$

$\Rightarrow\text{y}_1=-\text{a}\sin(\log_\text{e}\text{x})\frac{1}{\text{x}}+\text{b}\cos(\log_\text{e}\text{x})\frac{1}{\text{x}}$

$\Rightarrow\text{y}_2=\frac{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})}{\text{x}}$

$\Rightarrow\text{y}_2=\frac{-\text{a}(\log_\text{e}\text{x})-\text{b}\sin(\log_\text{e}\text{x})-\{-\text{a}\sin(\log_\text{e}\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}}{\text{x}^2}$

$\Rightarrow\text{x}^2\text{y}_2=-\{\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})\}\\-\{-\text{a}\sin(\log\text{x})+\text{b}\cos(\log_\text{e}\text{x})\}$

$\Rightarrow\text{x}^2\text{y}_2=-\text{y}-\text{xy}_1$

$\Rightarrow\text{x}^2\text{y}_2+\text{xy}_1=-\text{y}$