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NCERT Solutions · Class 12 Mathematics Differential Equations

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Miscellaneous Exercise 1–10 (part 9 of 10)

  1. Exercise 1

    For each of the differential equations given below, indicate its order and degree (if defined).
    (i)
    d2ydx2+5x(dydx)2−6y=log⁡x\displaystyle \frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x
    (ii)
    (dydx)3−4(dydx)2+7y=sin⁡x\displaystyle \left(\frac{d y}{d x}\right)^{3}-4\left(\frac{d y}{d x}\right)^{2}+7 y=\sin x
    (iii)
    d4ydx4−sin⁡(d3ydx3)=0\displaystyle \frac{d^{4} y}{d x^{4}}-\sin \left(\frac{d^{3} y}{d x^{3}}\right)=0

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    NCERT’s answer
    (i)
    Order $\displaystyle 2$; Degree $\displaystyle 1$ (ii) Order $\displaystyle 1$; Degree $\displaystyle 3$ (iii) Order $\displaystyle 4$; Degree not defined
    The order of a differential equation is the order of the highest-order derivative occurring in it. Its degree is the power to which that highest-order derivative is raised, after the equation has been written as a polynomial in all the derivatives; if the equation cannot be put in such a polynomial form, the degree is not defined.
    (i)
    \(\displaystyle \frac{d^{2}y}{dx^{2}}+5x\left(\frac{dy}{dx}\right)^{2}-6y=\log x\)
    The highest-order derivative present is \(\displaystyle \frac{d^{2}y}{dx^{2}}\), so the order is \(\displaystyle 2\). The equation is already a polynomial in \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{d^{2}y}{dx^{2}}\), and \(\displaystyle \frac{d^{2}y}{dx^{2}}\) occurs to the first power. The square sits on \(\displaystyle \frac{dy}{dx}\), which is not the highest-order derivative, so it plays no part in the degree; \(\displaystyle \log x\) involves \(\displaystyle x\) only and is irrelevant to both.
    Order \(\displaystyle 2\), degree \(\displaystyle 1\).
    (ii)
    \(\displaystyle \left(\frac{dy}{dx}\right)^{3}-4\left(\frac{dy}{dx}\right)^{2}+7y=\sin x\)
    The only derivative present is \(\displaystyle \frac{dy}{dx}\), so the order is \(\displaystyle 1\). The equation is a polynomial in \(\displaystyle \frac{dy}{dx}\), and here the highest-order derivative is itself raised to the power \(\displaystyle 3\).
    Order \(\displaystyle 1\), degree \(\displaystyle 3\).
    (iii)
    \(\displaystyle \frac{d^{4}y}{dx^{4}}-\sin\left(\frac{d^{3}y}{dx^{3}}\right)=0\)
    The highest-order derivative is \(\displaystyle \frac{d^{4}y}{dx^{4}}\), so the order is \(\displaystyle 4\). But \(\displaystyle \frac{d^{3}y}{dx^{3}}\) appears inside a sine; expanding \(\displaystyle \sin t=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\cdots\) with \(\displaystyle t=\frac{d^{3}y}{dx^{3}}\) gives an infinite series, so the equation can never be written as a polynomial in the derivatives.
    Order \(\displaystyle 4\), degree not defined.

  2. Exercise 2

    For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
    (i)
    xy=aex+be−x+x2:xdx2+2ddx−xy+x2−2=0\displaystyle x y=a e^{x}+b e^{-x}+x^{2} \quad: x \frac{}{d x^{2}}+2 \frac{d}{d x}-x y+x^{2}-2=0
    (ii)
    y=ex(acos⁡x+bsin⁡x):d2ydx2−2dydx+2y=0\displaystyle y=e^{x}(a \cos x+b \sin x) \quad: \quad \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0
    (iii)
    y=xsin⁡3x:dx2+9y−6cos⁡3x=0\displaystyle y=x \sin 3 x \quad: \frac{}{d x^{2}}+9 y-6 \cos 3 x=0
    (iv)
    x2=2y2log⁡y:(x2+y2)dydx−xy=0\displaystyle x^{2}=2 y^{2} \log y \quad:\left(x^{2}+y^{2}\right) \frac{d y}{d x}-x y=0

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    Reading note: the printed text here has lost the numerators of two fractions. The denominators fix them uniquely, so the equations to be verified are \(\displaystyle x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}-xy+x^{2}-2=0\) in (i) and \(\displaystyle \frac{d^{2}y}{dx^{2}}+9y-6\cos 3x=0\) in (iii); the verifications below close exactly, which confirms the reading.
    Method in each part: differentiate the given relation as many times as the order of the equation, then use the original relation to eliminate the arbitrary constants, and substitute into the left-hand side.
    (i)
    \(\displaystyle xy=ae^{x}+be^{-x}+x^{2}\).
    Differentiate with respect to \(\displaystyle x\), using the product rule on the left:
    \[x\frac{dy}{dx}+y=ae^{x}-be^{-x}+2x.\]
    Differentiate once more:
    \[x\frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}+\frac{dy}{dx}=ae^{x}+be^{-x}+2,\qquad\text{i.e.}\qquad x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=ae^{x}+be^{-x}+2.\]
    The step that matters is eliminating \(\displaystyle a,b\): the original relation gives \(\displaystyle ae^{x}+be^{-x}=xy-x^{2}\). Hence
    \[x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}=xy-x^{2}+2,\]
    that is \(\displaystyle x\frac{d^{2}y}{dx^{2}}+2\frac{dy}{dx}-xy+x^{2}-2=0\). The given function is a solution.
    (ii)
    \(\displaystyle y=e^{x}(a\cos x+b\sin x)\).
    \[\frac{dy}{dx}=e^{x}(a\cos x+b\sin x)+e^{x}(-a\sin x+b\cos x)=y+e^{x}(-a\sin x+b\cos x).\]
    So \(\displaystyle e^{x}(-a\sin x+b\cos x)=\frac{dy}{dx}-y\). Differentiating the displayed line again,
    \[\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}+\Big[e^{x}(-a\sin x+b\cos x)+e^{x}(-a\cos x-b\sin x)\Big]=\frac{dy}{dx}+\left(\frac{dy}{dx}-y\right)-y.\]
    Therefore \(\displaystyle \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-2y\), i.e. \(\displaystyle \frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+2y=0\). The given function is a solution.
    (iii)
    \(\displaystyle y=x\sin 3x\).
    \[\frac{dy}{dx}=\sin 3x+3x\cos 3x,\qquad \frac{d^{2}y}{dx^{2}}=3\cos 3x+3\cos 3x-9x\sin 3x=6\cos 3x-9x\sin 3x.\]
    Since \(\displaystyle x\sin 3x=y\), this is \(\displaystyle \frac{d^{2}y}{dx^{2}}=6\cos 3x-9y\), i.e. \(\displaystyle \frac{d^{2}y}{dx^{2}}+9y-6\cos 3x=0\). The given function is a solution.
    (iv)
    \(\displaystyle x^{2}=2y^{2}\log y\) (so \(\displaystyle y>0\), for \(\displaystyle \log y\) to exist).
    Differentiate implicitly with respect to \(\displaystyle x\), using the product rule on \(\displaystyle y^{2}\log y\):
    \[2x=2\left[2y\log y\cdot\frac{dy}{dx}+y^{2}\cdot\frac{1}{y}\cdot\frac{dy}{dx}\right]=2\frac{dy}{dx}\big(2y\log y+y\big),\]
    so \(\displaystyle x=\frac{dy}{dx}\big(2y\log y+y\big)\).
    Now eliminate \(\displaystyle \log y\): the given relation gives \(\displaystyle 2y\log y=\frac{x^{2}}{y}\). Hence
    \[x=\frac{dy}{dx}\left(\frac{x^{2}}{y}+y\right)=\frac{dy}{dx}\cdot\frac{x^{2}+y^{2}}{y},\]
    so \(\displaystyle xy=(x^{2}+y^{2})\frac{dy}{dx}\), i.e. \(\displaystyle (x^{2}+y^{2})\frac{dy}{dx}-xy=0\). The given implicit function is a solution.

  3. Exercise 3

    Prove that x2−y2=c(x2+y2)2\displaystyle x^{2}-y^{2}=c\left(x^{2}+y^{2}\right)^{2} is the general solution of differential equation (x3−3xy2)dx=(y3−3x2y)dy\displaystyle \left(x^{3}-3 x y^{2}\right) d x=\left(y^{3}-3 x^{2} y\right) d y, where c\displaystyle c is a parameter.

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    Write the equation as \[\frac{dy}{dx}=\frac{x^{3}-3xy^{2}}{y^{3}-3x^{2}y}.\] Numerator and denominator are both homogeneous of degree \(\displaystyle 3\), so the equation is homogeneous of degree zero and the substitution \(\displaystyle y=vx\), \(\displaystyle \frac{dy}{dx}=v+x\frac{dv}{dx}\) applies.Substituting, \[v+x\frac{dv}{dx}=\frac{x^{3}-3x\cdot v^{2}x^{2}}{v^{3}x^{3}-3x^{2}\cdot vx}=\frac{x^{3}(1-3v^{2})}{x^{3}(v^{3}-3v)}=\frac{1-3v^{2}}{v^{3}-3v}\qquad(x\neq 0).\] Hence \[x\frac{dv}{dx}=\frac{1-3v^{2}-v(v^{3}-3v)}{v^{3}-3v}=\frac{1-3v^{2}-v^{4}+3v^{2}}{v^{3}-3v}=\frac{1-v^{4}}{v^{3}-3v},\] and the variables separate: \[\frac{v^{3}-3v}{1-v^{4}}\,dv=\frac{dx}{x}.\]The left integral is the step to be careful with. The numerator is odd in \(\displaystyle v\), so put \(\displaystyle t=v^{2}\), \(\displaystyle dt=2v\,dv\), and use \(\displaystyle 1-v^{4}=(1-t)(1+t)\): \[\int\frac{v(v^{2}-3)}{(1-v^{2})(1+v^{2})}dv=\frac{1}{2}\int\frac{t-3}{(1-t)(1+t)}\,dt.\] Partial fractions: \(\displaystyle \dfrac{t-3}{(1-t)(1+t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}\) gives \(\displaystyle t-3=A(1+t)+B(1-t)\); putting \(\displaystyle t=1\) gives \(\displaystyle A=-1\) and \(\displaystyle t=-1\) gives \(\displaystyle B=-2\). So \[\frac{1}{2}\int\left(\frac{-1}{1-t}-\frac{2}{1+t}\right)dt=\frac{1}{2}\log|1-t|-\log|1+t|=\frac{1}{2}\log|1-v^{2}|-\log\left(1+v^{2}\right).\]Therefore \[\frac{1}{2}\log|1-v^{2}|-\log\left(1+v^{2}\right)=\log|x|+\log k.\] Multiply by \(\displaystyle 2\) and combine: \[\log\frac{|1-v^{2}|}{(1+v^{2})^{2}}=\log\left(Kx^{2}\right)\quad\Longrightarrow\quad \frac{1-v^{2}}{(1+v^{2})^{2}}=Kx^{2},\] the sign being absorbed into the arbitrary constant \(\displaystyle K\).Put back \(\displaystyle v=\dfrac{y}{x}\): \(\displaystyle 1-v^{2}=\dfrac{x^{2}-y^{2}}{x^{2}}\) and \(\displaystyle (1+v^{2})^{2}=\dfrac{(x^{2}+y^{2})^{2}}{x^{4}}\), so \[\frac{x^{2}(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}}=Kx^{2}\quad\Longrightarrow\quad x^{2}-y^{2}=K\left(x^{2}+y^{2}\right)^{2}.\]Writing \(\displaystyle K=c\), the relation \(\displaystyle x^{2}-y^{2}=c\left(x^{2}+y^{2}\right)^{2}\) satisfies the equation and contains exactly one arbitrary constant, which is the number required by a first-order equation. Hence it is the general solution.

  4. Exercise 4

    Find the general solution of the differential equation dydx+1−y21−x2=0\displaystyle \frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0.

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    NCERT’s answer
    \(\displaystyle \sin ^{-1} y+\sin ^{-1} x=\mathrm{C}\)
    Write the equation as \[\frac{dy}{dx}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}},\] which requires \(\displaystyle |x|<1\) and \(\displaystyle |y|\le 1\) for the square roots to be real with a non-zero denominator. The equation is then separable: \[\frac{dy}{\sqrt{1-y^{2}}}=-\frac{dx}{\sqrt{1-x^{2}}}.\]Integrating both sides, using \(\displaystyle \int\frac{dt}{\sqrt{1-t^{2}}}=\sin^{-1}t\), \[\sin^{-1}y=-\sin^{-1}x+\mathrm{C}.\]General solution: \(\displaystyle \sin^{-1}x+\sin^{-1}y=\mathrm{C}\), valid for \(\displaystyle -1<x<1,\ -1\le y\le 1\).

  5. Exercise 5

    Show that the general solution of the differential equation dydx+y2+y+1x2+x+1=0\displaystyle \frac{d y}{d x}+\frac{y^{2}+y+1}{x^{2}+x+1}=0 is given by (x+y+1)=A(1−x−y−2xy)\displaystyle (x+y+1)=\mathrm{A}(1-x-y-2 x y), where A is parameter.

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    The equation separates: \[\frac{dy}{y^{2}+y+1}+\frac{dx}{x^{2}+x+1}=0.\]Neither quadratic factorises over the reals, so complete the square: \(\displaystyle t^{2}+t+1=\left(t+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}\). Using \(\displaystyle \int\frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a}\), \[\int\frac{dt}{t^{2}+t+1}=\frac{2}{\sqrt{3}}\tan^{-1}\frac{2t+1}{\sqrt{3}}.\] Hence integrating the separated equation, \[\frac{2}{\sqrt{3}}\tan^{-1}\frac{2y+1}{\sqrt{3}}+\frac{2}{\sqrt{3}}\tan^{-1}\frac{2x+1}{\sqrt{3}}=\mathrm{C},\] so \[\tan^{-1}\frac{2x+1}{\sqrt{3}}+\tan^{-1}\frac{2y+1}{\sqrt{3}}=\mathrm{C}_{1},\qquad \mathrm{C}_{1}=\frac{\sqrt{3}}{2}\mathrm{C}.\]Now take the tangent of both sides, using \(\displaystyle \tan(\mathrm{A}+\mathrm{B})=\dfrac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A}\tan \mathrm{B}}\): \[\frac{\dfrac{2x+1}{\sqrt{3}}+\dfrac{2y+1}{\sqrt{3}}}{1-\dfrac{(2x+1)(2y+1)}{3}}=\tan \mathrm{C}_{1}.\] The numerator is \(\displaystyle \dfrac{2(x+y+1)}{\sqrt{3}}\) and the denominator is \[\frac{3-(4xy+2x+2y+1)}{3}=\frac{2\,(1-x-y-2xy)}{3}.\] Therefore \[\frac{2(x+y+1)}{\sqrt{3}}\cdot\frac{3}{2(1-x-y-2xy)}=\tan \mathrm{C}_{1}\quad\Longrightarrow\quad \frac{\sqrt{3}\,(x+y+1)}{1-x-y-2xy}=\tan \mathrm{C}_{1}.\]Writing \(\displaystyle \mathrm{A}=\dfrac{\tan \mathrm{C}_{1}}{\sqrt{3}}\), which is again an arbitrary parameter, \[(x+y+1)=\mathrm{A}\,(1-x-y-2xy),\] as required.

  6. Exercise 6

    Find the equation of the curve passing through the point (0,π4)\displaystyle \left(0, \frac{\pi}{4}\right) whose differential equation is sin⁡xcos⁡ydx+cos⁡xsin⁡ydy=0\displaystyle \sin x \cos y d x+\cos x \sin y d y=0.

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    NCERT’s answer
    \(\displaystyle \cos y=\frac{\sec x}{\sqrt{2}}\)
    The equation \(\displaystyle \sin x\cos y\,dx+\cos x\sin y\,dy=0\) is separable. Divide throughout by \(\displaystyle \cos x\cos y\) (assuming \(\displaystyle \cos x\neq 0,\ \cos y\neq 0\), which holds near the given point): \[\tan x\,dx+\tan y\,dy=0.\]Integrate, using \(\displaystyle \int\tan t\,dt=-\log|\cos t|\): \[-\log|\cos x|-\log|\cos y|=-\log k,\qquad\text{i.e.}\qquad \cos x\cos y=k.\]Apply the condition that the curve passes through \(\displaystyle \left(0,\frac{\pi}{4}\right)\): \[k=\cos 0\cdot\cos\frac{\pi}{4}=1\cdot\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}.\]Equation of the curve: \(\displaystyle \cos x\cos y=\dfrac{1}{\sqrt{2}}\), equivalently \(\displaystyle \sec x\sec y=\sqrt{2}\).

  7. Exercise 7

    Find the particular solution of the differential equation (1+e2x)dy+(1+y2)exdx=0\displaystyle \left(1+e^{2 x}\right) d y+\left(1+y^{2}\right) e^{x} d x=0, given that y=1\displaystyle y=1 when x=0\displaystyle x=0.

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    NCERT’s answer
    \(\displaystyle \tan ^{-1} y+\tan ^{-1}\left(e^{x}\right)=\frac{\pi}{2}\)
    Separate the variables in \(\displaystyle \left(1+e^{2x}\right)dy+\left(1+y^{2}\right)e^{x}dx=0\): \[\frac{dy}{1+y^{2}}=-\frac{e^{x}\,dx}{1+e^{2x}}.\]For the right side put \(\displaystyle t=e^{x}\), so \(\displaystyle dt=e^{x}dx\) and \(\displaystyle 1+e^{2x}=1+t^{2}\): \[\int\frac{e^{x}dx}{1+e^{2x}}=\int\frac{dt}{1+t^{2}}=\tan^{-1}t=\tan^{-1}\left(e^{x}\right).\]Integrating both sides therefore gives \[\tan^{-1}y=-\tan^{-1}\left(e^{x}\right)+\mathrm{C},\qquad\text{i.e.}\qquad \tan^{-1}y+\tan^{-1}\left(e^{x}\right)=\mathrm{C}.\]Use \(\displaystyle y=1\) when \(\displaystyle x=0\): \(\displaystyle \tan^{-1}1+\tan^{-1}1=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}\), so \(\displaystyle \mathrm{C}=\frac{\pi}{2}\).Particular solution: \(\displaystyle \tan^{-1}y+\tan^{-1}\left(e^{x}\right)=\dfrac{\pi}{2}\).

  8. Exercise 8

    Solve the differential equation yexydx=(xexy+y2)dy(y≠0)\displaystyle y e^{\frac{x}{y}} d x=\left(x e^{\frac{x}{y}}+y^{2}\right) d y(y \neq 0).

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    NCERT’s answer
    \(\displaystyle e^{\frac{x}{y}}=y+\mathrm{C}\)
    Write the equation as \[\frac{dx}{dy}=\frac{x\,e^{x/y}+y^{2}}{y\,e^{x/y}}\qquad(y\neq 0).\] Both sides are homogeneous of degree zero, and the exponent \(\displaystyle \frac{x}{y}\) points to treating \(\displaystyle x\) as the dependent variable. Put \[x=vy,\qquad \frac{dx}{dy}=v+y\frac{dv}{dy},\qquad e^{x/y}=e^{v}.\]Substituting, \[v+y\frac{dv}{dy}=\frac{vy\,e^{v}+y^{2}}{y\,e^{v}}=v+\frac{y}{e^{v}}.\] The \(\displaystyle v\) cancels from both sides — this is the step that makes the equation separable: \[y\frac{dv}{dy}=y\,e^{-v}\quad\Longrightarrow\quad e^{v}\,dv=dy.\]Integrate: \[e^{v}=y+\mathrm{C}.\]Restoring \(\displaystyle v=\dfrac{x}{y}\): \[e^{x/y}=y+\mathrm{C},\qquad y\neq 0,\] where \(\displaystyle \mathrm{C}\) is the constant of integration.

  9. Exercise 9

    Find a particular solution of the differential equation (x−y)(dx+dy)=dx−dy\displaystyle (x-y)(d x+d y)=d x-d y, given that y=−1\displaystyle y=-1, when x=0\displaystyle x=0. (Hint: put x−y=t\displaystyle x-y=t )

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    NCERT’s answer
    \(\displaystyle \log |x-y|=x+y+1\)
    Follow the hint and put \(\displaystyle t=x-y\). Then \(\displaystyle dt=dx-dy\), so \(\displaystyle dy=dx-dt\) and \[dx+dy=2\,dx-dt.\]The equation \(\displaystyle (x-y)(dx+dy)=dx-dy\) becomes \[t\,(2\,dx-dt)=dt\quad\Longrightarrow\quad 2t\,dx=(1+t)\,dt.\]Separating (\(\displaystyle t\neq 0\)): \[2\,dx=\frac{1+t}{t}\,dt=\left(\frac{1}{t}+1\right)dt.\] Integrate: \[2x=\log|t|+t+\mathrm{C}.\]Put \(\displaystyle t=x-y\) back: \[2x=\log|x-y|+(x-y)+\mathrm{C}\quad\Longrightarrow\quad x+y=\log|x-y|+\mathrm{C}.\]Apply \(\displaystyle y=-1\) when \(\displaystyle x=0\): \[0+(-1)=\log|0-(-1)|+\mathrm{C}=\log 1+\mathrm{C}=\mathrm{C},\] so \(\displaystyle \mathrm{C}=-1\).Particular solution: \(\displaystyle \log|x-y|=x+y+1\).

  10. Exercise 10

    Solve the differential equation [e−2xx−yx]dxdy=1(x≠0)\displaystyle \left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{d y}=1(x \neq 0).

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    NCERT’s answer
    \(\displaystyle y e^{2 \sqrt{x}}=(2 \sqrt{x}+\mathrm{C})\)
    Take the reciprocal of both sides of \(\displaystyle \left[\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}\right]\dfrac{dx}{dy}=1\): \[\frac{dy}{dx}=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\quad\Longrightarrow\quad \frac{dy}{dx}+\frac{1}{\sqrt{x}}\,y=\frac{e^{-2\sqrt{x}}}{\sqrt{x}}.\] This is a linear differential equation \(\displaystyle \frac{dy}{dx}+\mathrm{P}y=\mathrm{Q}\) with \(\displaystyle \mathrm{P}=x^{-1/2}\) and \(\displaystyle \mathrm{Q}=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}\) (take \(\displaystyle x>0\)).Integrating factor: \[\int \mathrm{P}\,dx=\int x^{-1/2}dx=2\sqrt{x},\qquad \text{I.F.}=e^{2\sqrt{x}}.\]The solution is \(\displaystyle y\cdot(\text{I.F.})=\int \mathrm{Q}\cdot(\text{I.F.})\,dx+\mathrm{C}\). The point of the problem is that the two exponentials cancel: \[y\,e^{2\sqrt{x}}=\int e^{2\sqrt{x}}\cdot\frac{e^{-2\sqrt{x}}}{\sqrt{x}}\,dx+\mathrm{C}=\int\frac{dx}{\sqrt{x}}+\mathrm{C}=2\sqrt{x}+\mathrm{C}.\]General solution: \(\displaystyle y\,e^{2\sqrt{x}}=2\sqrt{x}+\mathrm{C}\), i.e. \(\displaystyle y=\left(2\sqrt{x}+\mathrm{C}\right)e^{-2\sqrt{x}}\).