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I am stuck on the following problem that says:

Find the complete integral of $(p+q)(px+qy)=1$,where $p={ \partial z \over \partial x},q={ \partial z \over \partial y}$.

My Attempt:

The given equation is : $f(x,y,z,p,q)=(p+q)(px +qy)-1$. So,Charpit's auxiliary equations are given by: $${dp \over p(p+q)}={dq \over q(p+q)}= {dz \over 0}={dx \over {-q(x-y)}}={dy \over {-p(y-x)}}$$. Now, from $${dp \over p(p+q)}={dq \over q(p+q)} \implies {dp \over p}={dq \over q} \implies p=aq,\,a $$ being arbitray constant. Now, I have to use $$dz=pdx+qdy=q(adx+dy)$$ . Now,I am stuck. The answer is given of the form: $$z(a+1)^{\frac 12}=2(ax+b)^{\frac 12}+b$$

I have used the formula ${dz \over {-pf_p-qf_q}}={dp \over {f_x+pf_z}}={dq \over {f_y+qf_z}}={dx \over {-f_p}}={dy \over {-f_q}}$. Here, $f_x={\partial f \over \partial x}$,.....

Can someone help me out? Thanks in advance for your time.

learner
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  • Sorry, I should have mentioned. $p={ \partial z \over \partial x},q={ \partial z \over \partial y}$ – learner Sep 19 '13 at 14:37
  • I have used the formula ${dz \over {-pf_p-qf_q}}={dp \over {f_x+pf_z}}={dq \over {f_y+qf_z}}={dx \over {-f_p}}={dy \over {-f_q}}$. Here, $f_x={\partial f \over \partial x}$ – learner Sep 19 '13 at 14:52
  • i used the same formula for dz on mathematica, Simplify[- p D[(p + q) (p x + q y) - 1, {p, 1}] - q D[(p + q) (p x + q y) - 1, {q, 1}]] – S L Sep 19 '13 at 14:54
  • btw did you check that it satisfies the given PDE? – S L Sep 19 '13 at 15:03
  • check your charpit's equation – User8976 Mar 30 '15 at 02:23

1 Answers1

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The given equation is: $f(x,y,z,p,q)=(p+q)(px +qy)-1 = p^2x+pqy +pqx+q^2y-1 =0$

Thus the Charpit's auxiliary equations are: $${dp \over p(p+q)}={dq \over q(p+q)}= {dz \over ...}={dx \over {-(2px+qy+qx)}}={dy \over {-(py+px+2qy)}}$$ Taking the first two equations we get, $p =aq$ for some constant $a$. Putting it in original equation we get $$q^2 = {1 \over (a+1)(ax+y)}$$ and $$p^2 ={a^2 \over (a+1)(ax+y)}$$ Thus $p = {a \over \sqrt{(a+1)(ax+y)}}$ and $q= {1 \over \sqrt{(a+1)(ax+y)}}$. Putting this values of $p,q$ in $dz = pdx +qdy$ we get, $$dz = {d(ax+y)\over \sqrt{(a+1)(ax+y)}}$$

Then integrating we get, $$z(a+1)^{\frac 12}=2(ax+b)^{\frac 12}+b$$ where $a,b$ are constants.

User8976
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