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Consider the partial differential equation $$xp^2+yq^2+(x+y)(pq)-u(p+q)+1=0$$ where $p=\frac{\partial u}{\partial x}$ and $q=\frac{\partial u}{\partial y}$ Then which of the following statements are true ?

$1.$ The general solutions can be expressed as $u(x,y)=ax+by+\frac{1}{a+b}$ where $a$ and $b$ are arbitrary constants.

$2.$ The general solutions can be expressed as $u(x,y)=f(ax+by)+\frac{1}{a+b}$ where $a$ and $b$ are arbitrary constants and $f$ is arbitrary function.

$3.$ The Charpit equations are $$\frac{dx}{p^2+pq}=\frac{dy}{q^2+pq}=\frac{du}{p(p^2+pq)+q(q^2+pq)}=\frac{dp}{0}=\frac{dq}{0}$$

$4.$ The Charpit equations are $$\frac{dx}{2px+(x+y)q-u}=\frac{dy}{2qy+(x+y)p-u}=\frac{du}{p(2px+(x+y)q-u)+q(2qy+(x+y)p-u)}=\frac{dp}{0}=\frac{dq}{0}$$

I know that option three is wrong one and option 4 is correct one as Charpit auxiliary are are correct in Option $4$. Now in first two options it is asked about general solutions and Charpit method is to find complete integral . If we take $p=a$ and try to find $q$ so that complete integral is given by $du=pdx+qdy$, and then find general integral is not seems to be good in given time frame in exam . Please help me to solve first two options. Thanks you.

Gonçalo
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neelkanth
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1 Answers1

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It follows from Charpit's equations that both $p$ and $q$ are constants. Substituting $p=a$ and $q=b$ in $$ du=pdx+qdy \tag{1} $$ and integrating, we find $$ u=ax+by+c. \tag{2} $$ Finally, substituting $(2)$ in the PDE $$ xp^2+yq^2+(x+y)pq-u(p+q)+1=0 \tag{3} $$ and simplifying, we obtain $$ c=\frac{1}{a+b}. \tag{4} $$ Therefore, we conclude that statement $1$ is true (and $2$ is false).

Gonçalo
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