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Differential Equations Test - 40

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Differential Equations Test - 40
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  • Question 1
    1 / -0
    The differential equation representing a family of circles touching  the $$y$$-axis at the origin is 
    Solution
    $$(x-a)^2+y^2=a^2$$

    $$x^2+a^2-2ax+y^2=a^2$$

    $$x^2-2ax+y^2=0$$...(1)

    Differentiating w.r.t x

    $$2x-2a+2y\dfrac{dy}{dx}=0$$

    $$\Rightarrow a=x+y\dfrac{dy}{dx}$$

    Substituting the value of a in eqn (1), we get,

    $$x^2-2x\left ( x+y\dfrac{dy}{dx} \right )+y^2=0$$

    $$x^2-2x^2-2xy\dfrac{dy}{dx} +y^2=0$$

    $$y^2-x^2-2xy\dfrac{dy}{dx}=0$$

    $$\Rightarrow x^2-y^2+2xy\dfrac{dy}{dx}=0$$
  • Question 2
    1 / -0
    Solution of differential equation $$\dfrac{dy}{dx}-2xy=x$$ is
    Solution
    Given $$\dfrac{dy}{dx}-2xy=x$$ within in first order lines differential 

    equation of form $$\dfrac{dy}{dx}+ Py=Q$$

    $$I.F= e^{\int Pdx} = e^{\int (-2x)dx}= C^{-x^{2}}$$

    Multiplying $$I.F$$ on both sides 
    $$e^{-x^{2}} \dfrac{dy}{dx} - 2xy. e^{-x^{2} }= xe^{-x^{2}}$$ 
    $$d (e^{-x^{2} }.y) =xe^{-x^{2} }dx$$
    Integrating both sides 

    $$e^{-x^{2} }y= \int xe^{-x^{2} } dx$$
    $$=\dfrac{-e^{-x^{2}}}{2}+c$$

    $$\Rightarrow y= ce^{x^{2}}- \dfrac{1}{2}$$

    $$\therefore $$ Option $$A$$ is correct
  • Question 3
    1 / -0
    If $$2x = {y^{\dfrac{1}{5}}} + {y^{\dfrac{{ - 1}}{5}}}{\text{and}}\left( {{x^2} - 1} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + \lambda x\dfrac{{dy}}{{dx}} + {\text{ky}} = 0,\;{\text{then}}\;\lambda  + {\text{K}}$$ is equal to.
    Solution
    Given
    $$2x=y^{\dfrac{1}{5}}+y^{-\dfrac{1}{5}}$$
    or
    $$y^{\dfrac{1}{5}}+y^{-\dfrac{1}{5}}=2x$$
    Differentiate with respect to x on both the sides,
    $$\therefore \dfrac{1}{5}.y^{\dfrac{-4}{5}}\dfrac{dy}{dx}-\dfrac{1}{5}.y^{\dfrac{-6}{5}}.\dfrac{dy}{dx}=2$$

    $$\dfrac{y^{-1}}{5}(y^{\dfrac{1}{5}}-y^{\dfrac{-1}{5}})\dfrac{dy}{dx}=2$$

    $$(y^{\dfrac{1}{5}}-y^{-\dfrac{1}{5}})\dfrac{dy}{dx}=10y$$

    $$(2\sqrt{x^2-1})\dfrac{dy}{dx}=10y$$$$~~~~~~~-(1)$$

    Again Differentiating above equation with respect to x,$$
    $$2\sqrt{x^2-1}\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}.2.\dfrac{2x}{2\sqrt{x^2-1}}=10\dfrac{dy}{dx}$$
    $$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=5\sqrt{x^2-1}\dfrac{dy}{dx}$$
    $$(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}=5.5y$$   (  from equation (1))

    On comparing with the above equation,
    we get ,$$\lambda=1 and k=-25$$

    So $$\lambda+k=1-25=-24$$
  • Question 4
    1 / -0
    The solution of $$y d x - x d y + 3 x ^ { 2 } y ^ { 2 } e ^ { x ^ { 3 } } d x = 0$$
    Solution
    $$\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}+3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}=0$$
    Consider $$\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}$$
    We know that $$\dfrac{d}{dx}\left(\dfrac{x}{y}\right)=\dfrac{ydx-xdy}{{y}^{2}}$$
    $$\displaystyle\int{\dfrac{d}{dx}\left(\dfrac{x}{y}\right)dx}$$
    $$=\dfrac{x}{y}+{c}_{1}$$
    Consider $$3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}$$
    $$=\displaystyle\int{3{x}^{2}{e}^{{x}^{3}}dx}$$
    Let $$t={x}^{3}\Rightarrow\,dt=3{x}^{2}dx$$
    $$=\displaystyle\int{{e}^{t}dt}$$
    $$={e}^{t}+{c}_{2}$$
    $$={e}^{{x}^{3}}+{c}_{2}$$
    $$\therefore\,\displaystyle\int{\dfrac{ydx-xdy}{{y}^{2}}}+3\displaystyle\int{{x}^{2}{e}^{{x}^{3}}dx}=0$$
    $$\dfrac{x}{y}+{c}_{1}+{e}^{{x}^{3}}+{c}_{2}=0$$
    $$\Rightarrow\,\dfrac{x}{y}+{e}^{{x}^{3}}=-\left({c}_{1}+{c}_{2}\right)$$
    $$\Rightarrow\,\dfrac{x}{y}+{e}^{{x}^{3}}=c$$ where $$c=-\left({c}_{1}+{c}_{2}\right)$$
  • Question 5
    1 / -0
    The order of the differential equation of all circles having radius $$r$$ is.
    Solution
    The equation of family of circle is $$(x-a)^2+(y-b)^2=r^2$$               ---- ( 1 )
    Clearly, the given equation have two arbitrary constants, so we differentiate twice w.r.t $$x$$
    $$\Rightarrow$$  $$2(x-a)+2(y-b)=0$$

    $$\Rightarrow$$  $$(x-a)+(y-b)\dfrac{dy}{dx}=0$$ 

    $$\Rightarrow$$  $$(x-a)=-(y-b)\dfrac{dy}{dx}$$                    ---- ( 2 )

    Differentiate again w.r.t $$x,$$
    $$1+(y-b)\dfrac{d^2y}{dx^2}+\left(\dfrac{dy}{dx}\right)^2=0$$

    $$\Rightarrow$$  $$(y-b)=-\dfrac{1+\left(\dfrac{dy}{dx}\right)^2}{\dfrac{d^2y}{dx^2}}$$                 ---- ( 3 )

    Substitute ( 3 ) and ( 2 ) in ( 1 ), we get,

    $$\Rightarrow$$  $$\left[\dfrac{1+\left(\dfrac{dy}{dx}\right)^2}{\dfrac{d^2y}{dx^2}}\dfrac{dy}{dx}\right]^2$$$$+\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^2}{\dfrac{d^2y}{dx^2}}=r^2$$

    $$\Rightarrow$$  $$\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^2\left[1+\left(\dfrac{dy}{dx}\right)^2\right]=r^2\left(\dfrac{d^2y}{dx^2}\right)^2$$

    $$\Rightarrow$$  $$\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^3=r^2\left(\dfrac{d^2y}{dx^2}\right)^2$$

    $$\therefore$$  We can see required order is $$2$$
    as the Order of a differential equation is the order of the highest derivative present in the equation.
  • Question 6
    1 / -0
    The solution of, $$\dfrac{xdy}{x^2 + y^2} = \left(\dfrac{y}{x^2 + y^2} - 1 \right)dx$$, is given by
    Solution
    $$\dfrac{xdy}{x^2+y^2}=\left(\dfrac{y}{x^2+y^2}-1\right)dx$$
    $$\dfrac{xdy}{x^2+y^2}=\dfrac{ydx}{x^2+y^2}-dx$$
    $$\dfrac{xdy-ydx}{x^2+y^2}=-dx$$
    $$\dfrac{d}{dx}\tan^{-1}\dfrac{y}{x}=-dx$$
    Integrating both sides, we get
    $$\tan^{-1}\dfrac{y}{x}=-x+c$$
    $$\tan^{-1}\dfrac{y}{x}+x=c$$.
  • Question 7
    1 / -0
    The solution of the differential equation $$\operatorname { xdy } \left( y ^ { 2 } e ^ { x y } + e ^ { \tfrac { x } { y } } \right) = y d x \left( e ^ { \frac { x } { y } } - y ^ { 2 } e ^ { x y } \right)$$ is-
    Solution
    $$xdy(y^2e^{xy}+e^{x/y}=ydx(e^{x/y}-y^2e^{xy})$$
    $$\dfrac{dy}{dx}x(y^2e^{xy}+e^{x/y})=y(e^{x/y}-y^2e^{xy})$$
    $$y'(xy^2e^{xy}+xe^{x/y})=ye^{x/y}-y^3e^{xy}$$
    $$e^{xy}(xy^2y'+y^3)=(y-y'x)e^{x/y}$$
    $$e^{xy}(xy'+y)=\left(\dfrac{1}{y}-\dfrac{y'x}{y^2}\right)e^{x/y}$$
    (Dividing by $$y^2$$)
    $$\dfrac{d}{dx}(e^{xy})=\dfrac{d}{dx}(e^{x/y})$$
    $$\Rightarrow e^{xy}=e^{x/y}+c$$
    $$\Rightarrow xy=ln|e^{x/y}+c|$$.
  • Question 8
    1 / -0
    The solution of the differential equation $$( x \cot y + \log \cos x ) d y$$ $$+ ( \log \sin y - y \tan x ) d x = 0$$ is:-
    Solution

  • Question 9
    1 / -0
    The solution of the differential equation $$y ^ { 2 } d y = x ( y d x - x d y ) \text { is } y = y ( x )$$. If $$y ( \sqrt { 3 } e ) = e$$ and $$y \left( x _ { 0 } \right) = 1$$ then $$x_0$$ is
    Solution
    $${ y }^{ 2 }dy=x(ydx-xdy)$$

    $$1=\dfrac { x }{ y } \frac { dy }{ dx } -\dfrac { { x }^{ 2 } }{ { y }^{ 2 } } .....(1)$$

    let $$x=vy$$

    then, $$\dfrac { dx }{ dy } =v+y\dfrac { dv }{ dy }$$

    now putting this in equation (1), we get

    $$1+{ v }^{ 2 }=v(v+vy\dfrac { dv }{ dy } )$$

    $$ 1=vy \dfrac { dv }{ dy }$$

    integrating the above, we get

    $$\int { \dfrac { dy }{ y }  } =\int { vdv } = log y=\dfrac { { v }^{ 2 } }{ 2 } +C$$ 

    $$log y=\dfrac { { x }^{ 2 } }{ { 2y }^{ 2 } } +C$$

    Using the initial conditions, we get 

    $$1=\dfrac { 3 }{ 2 } +C\Longrightarrow C=\dfrac { -1 }{ 2 } $$

    $$log (y)=\dfrac { { x }^{ 2 } }{ { 2y }^{ 2 } } -\dfrac { 1 }{ 2 }$$ when $$ y=0,$$ then 

    $$log (1)=\dfrac { { x }^{ 2 } }{ { 2(1) }^{ 2 } } -\dfrac { 1 }{ 2 } $$ 

    $$0=\dfrac { { x }^{ 2 } }{ { 2 } } -\dfrac { 1 }{ 2 } \Longrightarrow  \dfrac { { x }^{ 2 } }{ { 2 } } =\dfrac { 1 }{ 2 }$$

    $$ { x }^{ 2 }=1\Longrightarrow { x }_{ 0 }=1 $$ (Answer) 
    So, option (B) is correct.
  • Question 10
    1 / -0
    The differential equation whose solution is $$y = Ax^{5} + Bx^{4}$$ is
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