Question

deferenciation eq ^n solve pls
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Answer 1

Question :-

Solve the differential equation

$\rm \: ( {x}^{2} + xy)dy \: = \: ( {x}^{2} + {y}^{2})dx$

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm \: ( {x}^{2} + xy)dy \: = \: ( {x}^{2} + {y}^{2})dx$

can be rewritten as

$\rm \: \dfrac{dy}{dx} = \dfrac{ {x}^{2} + {y}^{2} }{ {x}^{2} + xy}$

As degree of each term of numerator and denominator is same. So, its a homogeneous differential equation.

So, to solve such differential equation, we substitute

$\red{\rm \: y = vx - - - (1)}$

On differentiating both sides w. r. t. x, we get

$\red{\rm \: \dfrac{dy}{dx} = v\dfrac{d}{dx}x + x\dfrac{d}{dx}v}$

$\red{\rm \: \dfrac{dy}{dx} = v \times 1 + x\dfrac{dv}{dx}}$

$\red{\rm \: \dfrac{dy}{dx} = v + x\dfrac{dv}{dx}}$

So, on substituting these values, we get

$\rm \: v + x\dfrac{dy}{dx} = \dfrac{ {x}^{2} + {(vx)}^{2} }{ {x}^{2} + x(vx)}$

$\rm \: v + x\dfrac{dy}{dx} = \dfrac{ {x}^{2} + {v}^{2} {x}^{2} }{ {x}^{2} + {vx}^{2} }$

$\rm \: v + x\dfrac{dy}{dx} = \dfrac{ {x}^{2}(1 + {v}^{2})}{{x}^{2}(1 +v) }$

$\rm \: v + x\dfrac{dy}{dx} = \dfrac{ 1 + {v}^{2}}{1 +v }$

$\rm \: x\dfrac{dy}{dx} = \dfrac{ 1 + {v}^{2}}{1 +v } - v$

$\rm \: x\dfrac{dy}{dx} = \dfrac{ 1 + {v}^{2} - v - {v}^{2} }{1 +v }$

$\rm \: x\dfrac{dy}{dx} = \dfrac{ 1- v}{1 +v }$

On separating the variables, we get

$\rm \: \dfrac{dx}{x} = \dfrac{ 1 + v}{1 - v } \: dv$

On integrating both sides, we get

$\rm \: \displaystyle\int\rm \dfrac{dx}{x} = \displaystyle\int\rm \dfrac{ 1 + v}{1 - v } \: dv$

$\rm \: log x = - \displaystyle\int\rm \dfrac{v + 1}{v - 1} \: dv$

$\rm \: log x = - \displaystyle\int\rm \dfrac{v - 1 + 1+ 1}{v - 1} \: dv$

$\rm \: log x = - \displaystyle\int\rm \dfrac{v - 1 +2}{v - 1} \: dv$

$\rm \: log x = - \displaystyle\int\rm \bigg(1 + \dfrac{2}{v - 1} \bigg)\: dv$

$\rm \: log x = - (v + 2log(v - 1)) + c$

$\rm \: log x = - v - 2log(v - 1) + c$

$\rm \: log x + v + 2log(v - 1) = c$

$\rm \: log x + \frac{y}{x} + 2log\bigg(\dfrac{y}{x} - 1\bigg) = c$

$\rm \: log x + \frac{y}{x} + 2log\bigg(\dfrac{y - x}{x}\bigg) = c$

$\rm \: logx+ \frac{y}{x} + 2[log(y - x) - logx]= c$

$\rm \: logx+ \frac{y}{x} + 2log(y - x) - 2logx= c$

$\rm\implies \:\rm \: \frac{y}{x} + 2log(y - x) - logx= c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

Sol.

The given equation is

$\\ \pmb{\[ \begin{array}{l}\displaystyle\tt \left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x \\ \\ \displaystyle\tt \therefore \quad \frac{d y}{d x}=\frac{x^{2}+y^{2}}{x^{2}+x y} \qquad \ldots(i) \end{array} \]}$

which is a homogeneous differential equation. Put y=vx

$\text{so that \( \tt \dfrac{d y}{d x}=v+x \dfrac{d v}{d x} \) Putting value of \( \tt\dfrac{d y}{d x} \) and \( \tt y \) in (i),}$

we have

$\begin{aligned}& \tt v+x \frac{d v}{d x} =\frac{x^{2}+v^{2} x^{2}}{x^{2}+v x^{2}} \\ \\ \Rightarrow & \tt v+x \frac{d v}{d x}= \frac{x^{2}\left(1+v^{2}\right)}{x^{2}(1+v)} \\ \\ \Rightarrow & \tt x \frac{d v}{d x}=\frac{1+v^{2}}{1+v}-v \\ \\ \Rightarrow & \tt x \frac{d v}{d x}=\frac{1+v^{2}-v-v^{2}}{1+v} \\ \\ \Rightarrow & \tt \frac{1+v}{1-v} d v=\frac{1}{x} d x \end{aligned}$

Integrating both sides, we have

$\[ \begin{aligned} & \tt \frac{1+v}{1-v} d v=\int \frac{1}{x} d x \Rightarrow \int\left(-1+\frac{2}{1-v}\right) d v \\ \\ & \tt \therefore-v-2 \log |1-v|=\log |x|-\log c \\ \\ \Rightarrow & \tt-\frac{y}{x}-2 \log \left|1-\frac{y}{x}\right|=\log |x|-\log c \\ \\ \Rightarrow & \tt-\frac{y}{x}-\log \frac{(x-y)^{2}}{x^{2}}=\log \left|\frac{x}{c}\right| \\ \\ \Rightarrow & \tt-\frac{y}{x}=\log \frac{(x-y)^{2}}{x^{2}} \times \frac{x}{c} \\ \\ \Rightarrow & \tt \frac{(x-y)^{2}}{c x}=e^{-y / x} \\ \\ \Rightarrow& \tt(x-y)^{2}=c x e^{-y / x} \end{aligned} \]$

which is required solution.