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Need help solving this.

Find the particular solution of the differential equation $$u_y = (5x + 2)u$$ that satisfies the data $u(x, x^2) = x^3$.

I usually try to find the characteristic equation but I can only see $u_y$ here.

J. W. Tanner
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hadams12
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  • How is this a differential equation? I don’t see any differentials here. I’m confused. – Radial Arm Saw Aug 18 '20 at 13:55
  • It is a partial differential equation – hadams12 Aug 18 '20 at 13:57
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    @RadialArmSaw The subscript notation for differentials is particularly aggravating to me as well. – K.defaoite Aug 18 '20 at 14:05
  • @K.defaoite It is weird. I’ve never ever seen differentials or partial differentials denoted that way. – Radial Arm Saw Aug 18 '20 at 14:06
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    I have seen partial differentials written this way; to me the notation makes perfect sense but for the avoidance of doubt I would write the dependent variables for $u$, as as far as we know $u$ could be a function of $4$ vaiables $u(x, y, z, w)$ (I am sure it is just $x$ and $y$). –  Aug 18 '20 at 14:08
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    @RadialArmSaw Frustratingly, although very misleading, the notation is commonly used. Personally I prefer $\partial_x^n$ representing $\frac{\partial^n}{\partial x^n}$ and likewise $\mathrm{D}_x^n$ representing $\frac{\mathrm{d}^n}{\mathrm{d}x^n}$. – K.defaoite Aug 18 '20 at 14:10

1 Answers1

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We treat is as an ODE, using a suitable integral factor: $$ u_y(x,y)=(5x+2)u(x,y)\quad\Longleftrightarrow\quad \mathrm{e}^{-(5x+2)y} u_y(x,y)-\mathrm{e}^{-(5x+2)y}(5x+2)u(x,y)=0 \\ \quad\Longleftrightarrow\quad\frac{\partial}{\partial y}\left(\mathrm{e}^{-(5x+2)y}u(x,y)\right)=0 \quad\Longleftrightarrow\quad \mathrm{e}^{-(5x+2)y}u(x,y)=f(x) $$ for some function $f(x)$ to be found.

Hence $$ u(x,y)=\mathrm{e}^{(5x+2)y}f(x). $$ Now, $u(x,x^2)=x^3$, provides that $$ x^3=u(x,x^2)=\mathrm{e}^{(5x+2)x^2}f(x) $$ and hence $$ f(x)=\mathrm{e}^{-(5x+2)x^2}x^3 $$ Altogether $$ u(x,y)=\mathrm{e}^{(5x+2)y}f(x)=\mathrm{e}^{(5x+2)y}\mathrm{e}^{-(5x+2)x^2}x^3= \mathrm{e}^{(5x+2)(y-x^2)}x^3. $$

Yiorgos S. Smyrlis
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