which ode does y=sin(log(x)) satisfy?

Question : Which ode does y=sin(log(x)) satisfy? Solution : Given function \[y=sin\left(log\left(x\right)\right)\] differentiating with respect to x \[\frac{dy}{dx}=\cos{\left(\log{\left(x\right)}\right)}\frac{1}{x}\] \[x\ \frac{dy}{dx}=\cos{\left(\log{\left(x\right)}\right)}\] Again differentiating with respect to x \[\frac{dy}{dx}+x\frac{d^2y}{dx^2}=-sin\left(log\left(x\right)\right)\frac{1}{x}=\frac{y}{x}\] \[x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0\] is the…

How to prove that F=(x^2-y^2)i-(2xy+y)j is a conservative force

Question : Prove that F=(x^2-y^2)i-(2xy+y)j is a conservative force Solution : To prove that Force\[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] is a conservative force. We shall find curl of F \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial…

how to Find curl of the vector F=(x^2-y^2)i-(2xy+y)j

Question : Find curl of the vector F=(x^2-y^2)i-(2xy+y)j Solution : Given vector \[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] Then curl of F is \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial x}-\left(\hat{i}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial x}+\left(\hat{j}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial y}-\left(\hat{j}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial y}\] Since \[\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=0\ ,\…

Solved find the sum of the series (1/n+1/(n+1)+1/(n+2)+…..+1/4n as limit n approaches infinity

Question : Find the sum of the series 1/(n+1)+1/(n+2)+…..+1/4n as limit n approaches infinity Solution : Let the sum of the series as \[L=\lim_{n\rightarrow\infty}{\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots..+\frac{1}{4n}\right]\ }\] \[L=\lim_{n\rightarrow\infty}{\left[\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\ldots..+\frac{1}{n+3n}\right]\ }\] \[=\lim_{n\rightarrow\infty}{\frac{1}{n}\left[\frac{n}{n}+\frac{n}{n+1}+\frac{n}{n+2}+\ldots..+\frac{n}{n+3n}\right]\ }\] \[=\lim_{n\rightarrow\infty}{\frac{1}{n}\left[\frac{1}{1+\frac{0}{n}}+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}+\ldots..+\frac{1}{1+\frac{3n}{n}}\right]\…