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Question Number 141303 by BHOOPENDRA last updated on 17/May/21

Commented by BHOOPENDRA last updated on 17/May/21

$$iv$$

Answered by Dwaipayan Shikari last updated on 17/May/21

$$\mathbf{M}=\left(\begin{array}{c}\frac{1-cos\theta }{2}\frac{1}{4}{sin}^2\theta \\ 1\frac{1+cos\theta }{2}\end{array}\right)$$$${\mathbf{M}}^2=\left(\begin{array}{c}{sin}^2\frac{\theta }{2}{sin}^2\frac{\theta }{2}{cos}^2\frac{\theta }{2}\\ 1{cos}^2\frac{\theta }{2}\end{array}\right)\left(\begin{array}{c}{sin}^2\frac{\theta }{2}{sin}^2\frac{\theta }{2}{cos}^2\frac{\theta }{2}\\ 1{cos}^2\frac{\theta }{2}\end{array}\right)$$$$=\left(\begin{array}{c}{sin}^2\frac{\theta }{2}{sin}^2\frac{\theta }{2}{cos}^2\frac{\theta }{2}\\ 1{cos}^2\frac{\theta }{2}\end{array}\right)$$$${\mathbf{M}}^n=\left(\begin{array}{c}{sin}^2\frac{\theta }{2}{sin}^2\frac{\theta }{2}{cos}^2\frac{\theta }{2}\\ 1{cos}^2\frac{\theta }{2}\end{array}\right)$$

Answered by Dwaipayan Shikari last updated on 17/May/21

$${xy}^{\prime }-xy=y$$$$\Rightarrow {xy}^{\prime }=(x+1)y$$$$\Rightarrow \frac{dy}{dx}=\frac{x+1}{x}y\Rightarrow \int \frac{dy}{y}=\int \frac{x+1}{x}dx\Rightarrow log\left(y\right)=x+log\left(x\right)+C$$$$\Rightarrow log\left(\frac{y}{{xC}_1}\right)=x\Rightarrow y=C_1{xe}^x$$

Commented by BHOOPENDRA last updated on 17/May/21

$$tqsirbutinupperproblemialsoneed$$$$parttwo$$

Answered by Dwaipayan Shikari last updated on 17/May/21

$$x^2{y}^{\prime }+3xy=1$$$$\Rightarrow \frac{dy}{dx}+\frac{3y}{x}=\frac{1}{x^2}$$$$\mathrm{I}.\mathrm{F}=e^{\int \frac{3}{x}dx}=x^3$$$$x^{3}y=\int xdx=\frac{x^2}{2}+C\Rightarrow 2x^{3}y=x^2+C_1$$

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