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Differential Equations Test - 46

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Differential Equations Test - 46
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  • Question 1
    1 / -0
    The solution of $$ \dfrac{dv}{dt} + \dfrac{k}{m}v = -g $$ is
    Solution
    $$ \dfrac{dv}{dt} + \dfrac{k}{m} v = -g $$

    $$ \implies \dfrac{dv}{dt} = \dfrac{k}{m} (v + \dfrac{mg}{k}) $$

    $$ \implies \dfrac{dv}{v+mg/k} = \dfrac{k}{m} dt $$

    $$ \implies log (v + \dfrac{mg}{k}) = -\dfrac{k}{m}t + logc $$

    $$ \implies v + \dfrac{mg}{k} = c^{-(k/m)t} $$

    $$ \implies v = ce^{\dfrac{k}{m}t} - \dfrac{mg}{k} $$
  • Question 2
    1 / -0
    The solution of the differential equation $$ x^{2}\dfrac{dy}{dx}\cos \dfrac{1}{x} - y\sin\dfrac{1}{x} = -1, $$ where $$ y \rightarrow -1 $$ as $$ x \rightarrow \infty $$ is
    Solution
    $$ x^{2}\dfrac{dy}{dx}\cos\dfrac{1}{x}-y\sin\dfrac{1}{x}=-1 $$
    $$ \implies \dfrac{dy}{dx} - \dfrac{y}{x^{2}}\tan\dfrac{1}{x} =-\sec\dfrac{1}{x}\dfrac{1}{x^{2}} $$ (linear)
    I.F. $$ = e^{\int\dfrac{1}{x^{2}}\tan\dfrac{1}{x}dx}=\sec\dfrac{1}{x} $$
    $$ \implies $$ solutions is $$ y \sec\dfrac{1}{x} = -\int\sec^{2}(\dfrac{1}{x})\dfrac{1}{x^{2}}dx=\tan\dfrac{1}{x}+c $$
    Given $$ y \rightarrow -1, x \rightarrow \infty \implies x=-1 $$
    Hence equation of curve is $$ y=\sin\dfrac{1}{x} - \cos\dfrac{1}{x} $$
  • Question 3
    1 / -0
    The solution of $$ ye^{-x/y}dx-(xe^{(-x/y)} + y^{3}) dy=0 $$ is
    Solution
    $$ ye^{-x/y}dx-(xe^{-x/y} + y^{3})dy = 0 $$
    $$ \implies (ydx-xdy)e^{-x/y} - y^{3}dy = 0 $$
    $$ \implies \dfrac{ydx-xdy}{y^{2}}e^{-x/y} = ydy $$
    $$ \implies d(x/y)e^{-x/y} = ydy $$
    $$ -e^{-x/y} = \dfrac{y^{2}}{2} + C $$
    $$ 2 e^{-x/y}+ y^{2} = C $$
  • Question 4
    1 / -0
    Solution of the differential equation $$ (y+x\sqrt{xy}(x+y))dx + (y\sqrt{xy}(x+y)-x)dy = 0 $$ is
    Solution
    The given equation is written as $$ ydx-xdy+x\sqrt{xy}(x+y)dx+y\sqrt{xy}(x+y)dy=0 $$
    $$ \implies ydx-xdy+(x+y)\sqrt{xy}(xdx+ydy)=0 $$
    $$ \implies \dfrac{ydx-xdy}{y^{2}}+(\dfrac{x}{y}+1)\sqrt{x}{y}(d(\dfrac{x^{2}+y^{2}}{2}))=0 $$
    $$ \implies d\dfrac{x^{2}+y^{2}}{2}+\dfrac{d(\dfrac{x}{y})}{(\dfrac{x}{y} +1)\sqrt{\dfrac{x}{y}}} =0 $$
    $$ \implies \dfrac{x^{2}+y^{2}}{2}+2tan^{-1}\sqrt{\dfrac{x}{y}} = c $$
  • Question 5
    1 / -0
    Solution of differential equation $$ dy - \sin x \sin y dx = 0 $$ is
    Solution
    $$ dy - \sin x \sin y dx = 0 $$
    $$ \implies\ dy = \sin x \sin y dx  $$
    $$ \implies cosec\ y\ dy = \sin x  dx $$
    integating both side
    $$\implies log\tan\dfrac{y}{2}+log k=-\cos x$$
    $$\implies \dfrac{\tan\dfrac{y}{2}}{c}=e^{-\cos x}$$
    $$\implies e^{\cos x}\tan\dfrac{y}{2}=c$$
  • Question 6
    1 / -0
    The solution of, $$ydx-xdy+(1+x^2)dx+x^2 \sin y dy=0$$,
    Solution

  • Question 7
    1 / -0
    The solution of differential equation $$ \dfrac{x+y\dfrac{dy}{dx}}{y-x\dfrac{dy}{dx}} = \dfrac{x\cos^{2}(x^{2} + y^{2})}{y^{3}} $$ is
    Solution
    The given equation can be written as $$ \dfrac{xdx+ydy}{(ydx-xdy)/y^{2}}=y^{2}\dfrac{x}{y^{3}}\cos^{2}(x^{2}+y^{2}) $$

    $$ \implies \dfrac{xdx+ydy}{\cos^{2}(x^{2}+y^{2})}=\dfrac{x}{y}(\dfrac{ydx-xdy}{y^{2}}) $$

    $$ \implies \dfrac{1}{2}\sec^{2}(x^{2}+y^{2})d(x^{2}+y^{2})=\dfrac{x}{y}d(\dfrac{x}{y}) $$
    On integrating, we get
    $$\dfrac{1}{2}\tan(x^{2}+y^{2})=\dfrac{1}{2}(\dfrac{x}{y})^{2} + \dfrac{c}{2} $$
    or $$ \tan (x^{2}+y^{2})=\dfrac{x^{2}}{y^{2}}+ c $$
  • Question 8
    1 / -0
    The solution of $$\dfrac{dy}{dx}+x\sin 2y=x^{3}\cos^{2}y$$, is
    Solution

  • Question 9
    1 / -0
    Differential equation $$\dfrac{dy}{dx}=f(x)g(x)$$ can be solved by
    separating variable $$\dfrac{dy}{g(y)}=f(x)dx$$

    The equation of the curve to the point $$(1,0)$$ which
    stratifies the differential equation $$(1+y^2)dx-xydy=0$$
    Solution
    $$\dfrac{dx}{x}=\dfrac{ydy}{1+y^2}$$

    $$\Rightarrow  \ln$$ $$x=\dfrac{1}{2}.\ln\ (1+y^2)+\dfrac C2$$

    $$\Rightarrow  2\ln$$ $$x=\ln\ (1+y^2)+ C$$

    $$\Rightarrow  \ln$$ $$x^2=\ln\ (1+y^2)+ C$$

    From the given condition, $$(x,y)=(1,0),$$  $$C=0$$

    $$\Rightarrow  \ln$$ $$x^2=\ln\ (1+y^2)$$

    $$\Rightarrow x^2=1+y^2$$

    $$ \therefore x^2-y^2=1$$
  • Question 10
    1 / -0
    The solution of the differential equation $$(e^{x^2}+e^{y^2})y\dfrac{dy}{dx}+e^{x^{2}}(xy^2-x)=0$$ is
    Solution
    $$ y^2=t; 2y\dfrac{dy}{dx}; $$ hence the differential equation becomes
    $$ \left(e^{x^2}+e^t\right) +2e^{x62}(xt-x)=0 $$
    $$e^{x^2}+e^t+2e^{x^2}x(t-1)\dfrac{dx}{dt}=0$$
    put $$ e^{x^2}=z; e^{x^2}2x\dfrac{dx}{dt}=\dfrac{dz}{dt} $$
    $$ \implies \dfrac{dz}{dt}+\dfrac{z}{9t-1)}=\dfrac{e^t}{(t-1)}; $$ I.F. $$=e^{\int \dfrac{dt}{t-1}}=e^{\ln (t-1)} =t-1 $$
    $$ \implies z(t-1)=-\int(e^t)dt $$
    $$ \implies z(t-1)=e^t+C $$
    $$ \implies e^{x^2}(y^2-1)=-e^{y^2}+C$$
    $$ \implies e^{x^2}(y^2-1) + e^{y^2}=C$$
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