How to solve this PDE using Charpit's Method?
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If $p=fracpartial zpartial x$ and $q=fracpartial zpartial y$ then consider the following PDE:
$$p^2+q=0.$$
The using Charpit's Method we get the following auxiliary equations:
$$fracdxdt =2p$$
$$ fracdydt = 1 $$
$$ fracdzdt = 2p^2+q = p^2$$
$$ fracdpdt = 0$$
$$fracdqdt = 0.$$
Thus $p= c_1$ and $q= c_2$ with $c_2^2+c_1 = 0.$ Then
$$ x= 2c_1t+a_1 $$
$$y = t+a_2$$
$$z = tc_1^2+a_3.$$
Now I have the following initial conditions $z(x,0)=xd.$ How do I get the final solution? In particular, I am not sure how to eliminate $t$ from the three equations.
linear-pde
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up vote
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down vote
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If $p=fracpartial zpartial x$ and $q=fracpartial zpartial y$ then consider the following PDE:
$$p^2+q=0.$$
The using Charpit's Method we get the following auxiliary equations:
$$fracdxdt =2p$$
$$ fracdydt = 1 $$
$$ fracdzdt = 2p^2+q = p^2$$
$$ fracdpdt = 0$$
$$fracdqdt = 0.$$
Thus $p= c_1$ and $q= c_2$ with $c_2^2+c_1 = 0.$ Then
$$ x= 2c_1t+a_1 $$
$$y = t+a_2$$
$$z = tc_1^2+a_3.$$
Now I have the following initial conditions $z(x,0)=xd.$ How do I get the final solution? In particular, I am not sure how to eliminate $t$ from the three equations.
linear-pde
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $p=fracpartial zpartial x$ and $q=fracpartial zpartial y$ then consider the following PDE:
$$p^2+q=0.$$
The using Charpit's Method we get the following auxiliary equations:
$$fracdxdt =2p$$
$$ fracdydt = 1 $$
$$ fracdzdt = 2p^2+q = p^2$$
$$ fracdpdt = 0$$
$$fracdqdt = 0.$$
Thus $p= c_1$ and $q= c_2$ with $c_2^2+c_1 = 0.$ Then
$$ x= 2c_1t+a_1 $$
$$y = t+a_2$$
$$z = tc_1^2+a_3.$$
Now I have the following initial conditions $z(x,0)=xd.$ How do I get the final solution? In particular, I am not sure how to eliminate $t$ from the three equations.
linear-pde
If $p=fracpartial zpartial x$ and $q=fracpartial zpartial y$ then consider the following PDE:
$$p^2+q=0.$$
The using Charpit's Method we get the following auxiliary equations:
$$fracdxdt =2p$$
$$ fracdydt = 1 $$
$$ fracdzdt = 2p^2+q = p^2$$
$$ fracdpdt = 0$$
$$fracdqdt = 0.$$
Thus $p= c_1$ and $q= c_2$ with $c_2^2+c_1 = 0.$ Then
$$ x= 2c_1t+a_1 $$
$$y = t+a_2$$
$$z = tc_1^2+a_3.$$
Now I have the following initial conditions $z(x,0)=xd.$ How do I get the final solution? In particular, I am not sure how to eliminate $t$ from the three equations.
linear-pde
linear-pde
asked Aug 30 at 3:49
Hello_World
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3,20321429
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