Question
Solve the following differential equation:
$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=8\text{x}^2-3\text{xy}+2\text{y}^2$
$(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=8\text{x}^2-3\text{xy}+2\text{y}^2$
Integrating both sides, we get $\int\frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}\text{dv}=\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$ Let us consider the left hand side of (1). Using partial fraction, Let $\frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}=\frac{\text{Av + B}}{4+\text{v}^2}+\frac{\text{C}}{2-\text{v}}$ $\Rightarrow\ 1+\text{v}^2=\text{Av}(2-\text{v})+\text{B}(2-\text{v})+\text{C}(4+\text{v}^2)$ $\Rightarrow\ 1+\text{v}^2=2\text{Av}-\text{Av}^2+2\text{B}-\text{Bv}+4\text{C}+\text{Cv}^2$ Comparing the co-efficients of both sides, we get 2A - B = 0 -A + C = 1 & 2B + 4C = 1 Solving these three equations, we get $\text{A}=\frac{-3}8,\ \text{B}=\frac{-3}4$ and $\text{C}=\frac{5}8$ $\therefore\ \frac{1+\text{v}^2}{(4+\text{v}^2)(2-\text{v})}=\frac{-\frac{3}{8}\text{v}-\frac{3}{4}}{4+\text{v}^2}+\frac{\frac{5}{8}}{2-\text{v}}\ \dots(2)$ From (1) and (2), we get $\int\frac{-\frac{3}{8}\text{v}-\frac{3}{4}}{4+\text{v}^2}+\frac{\frac{5}{8}}{2-\text{v}}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ -\frac{3}8\int\frac{\text{v}}{\text{v}^2+4}\text{dv}-\frac{3}4\int\frac{1}{\text{v}^2+4}\text{dv}+\frac{5}8\int\frac{1}{2-\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \frac{-3}{16}\log|\text{v}^2+4|-\frac{3}{4\times2}\tan^{-1}\frac{\text{v}}2-\frac{5}8\log|2-\text{v}|=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ -\frac{3}{4\times2}\tan^{-1}\frac{\text{v}}2=\log\bigg|\text{Cx}(2-\text{v})^{\frac{5}8}(\text{v}^2+4)^{\frac{3}{16}}\bigg|$ $\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{v}}2}=\text{C}\bigg|\text{x}(2-\text{v})^{\frac{5}8}(\text{v}^2+4)^{\frac{3}{16}}\bigg|$ Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get $\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}\Bigg|\text{x}\Big(2-\frac{\text{y}}{\text{x}}\Big)^{\frac{5}8}\Big(\frac{\text{y}^2}{\text{x}^2}+4\Big)^{\frac{3}{16}}\Bigg|$ $\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}\bigg|\text{x}\times\frac{1}{\text{x}}(2\text{x}-\text{y})^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}\bigg|$ $\Rightarrow\ \text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}|2\text{x}-\text{y}|^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}$ Hence, $\text{e}^{-\frac{3}8\tan^{-1}\frac{\text{y}}{2\text{x}}}=\text{C}|2\text{x}-\text{y}|^{\frac{5}8}(\text{y}^2+4\text{x}^2)^{\frac{3}{16}}$ is the required solution.
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| Box | Marble colour | ||
|
| Red | White | Black |
| A B C D | 1 6 8 0 | 6 2 1 6 | 3 2 1 4 |
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?
$\int\sqrt{\tan\theta}\text{ d}\theta. $