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Question: How do I solve this partial differential equation \({{p}^{2}}={{z}^{2}}\left( 1-pq \right)\) ?...

How do I solve this partial differential equation
p2=z2(1−pq){{p}^{2}}={{z}^{2}}\left( 1-pq \right) ?

Explanation

Solution

To solve this question first we need to know whether this equation is a linear differential equation or nonlinear differential equation. The partial differential equation used here is a non linear differential equation which is not a function of xx and yy i.e. it’s only a function of p,q&z.p,q\And z.
To solve this, we will first assume as a function of which is a function of and then we will substitute the value of p&qp\And q where p=dzdup=\dfrac{dz}{du} and q=adzduq=a\dfrac{dz}{du} in the given equation and then we will solve the resulting ordinary differential equation in zz and uu
Then we will replace the value of u which we assumed as a function of x and y.

Complete step by step solution:
Let’s first assume the value of uu which is a function of x&yx\And y
u=x+ayu=x+ay
Also, zz is a function of u that is;
⇒z=f(u)\Rightarrow z=f\left( u \right)
We know,
⇒p=dzdu\Rightarrow p=\dfrac{dz}{du}& q=adzduq=a\dfrac{dz}{du}
Let’s put all the values in the given equation:-
⇒p2=z2(1−pq)\Rightarrow {{p}^{2}}={{z}^{2}}\left( 1-pq \right)
⇒(dzdu)2=z2[1−(dzdu)×a(dzdu)]\Rightarrow {{\left( \dfrac{dz}{du} \right)}^{2}}={{z}^{2}}\left[ 1-\left( \dfrac{dz}{du} \right)\times a\left( \dfrac{dz}{du} \right) \right]
Simplifying the equation further;
⇒(dzdu)2=z2[1−a(dzdu)2]\Rightarrow {{\left( \dfrac{dz}{du} \right)}^{2}}={{z}^{2}}\left[ 1-a{{\left( \dfrac{dz}{du} \right)}^{2}} \right]
⇒z2=(1−az2)(dydu)2\Rightarrow {{z}^{2}}=\left( 1-a{{z}^{2}} \right){{\left( \dfrac{dy}{du} \right)}^{2}}
Now, taking similar terms together;
⇒z2(1+az2)=(dzdu)2\Rightarrow \dfrac{{{z}^{2}}}{\left( 1+a{{z}^{2}} \right)}={{\left( \dfrac{dz}{du} \right)}^{2}}
Taking square root on both sides, we get;
⇒±z(1+az2)=(dzdu)\Rightarrow \pm \dfrac{z}{\sqrt{\left( 1+a{{z}^{2}} \right)}}=\left( \dfrac{dz}{du} \right)
Simplifying the equation further;
⇒±du=(1+az2)zdz\Rightarrow \pm du=\dfrac{\sqrt{\left( 1+a{{z}^{2}} \right)}}{z}dz
Multiplying numerator and denominator of right hand side by 1+az2\sqrt{1+a{{z}^{2}}}
⇒±du=(1+az2)×(1+az2)z×(1+az2)dz\Rightarrow \pm du=\dfrac{\sqrt{\left( 1+a{{z}^{2}} \right)}\times \sqrt{\left( 1+a{{z}^{2}} \right)}}{z\times \sqrt{\left( 1+a{{z}^{2}} \right)}}dz
⇒±du=1+az2z×(1+az2)dz\Rightarrow \pm du=\dfrac{1+a{{z}^{2}}}{z\times \sqrt{\left( 1+a{{z}^{2}} \right)}}dz
Breaking the terms of right hand side we get;
⇒±du=dzz×(1+az2)+azdzz×(1+az2)\Rightarrow \pm du=\dfrac{dz}{z\times \sqrt{\left( 1+a{{z}^{2}} \right)}}+\dfrac{azdz}{z\times \sqrt{\left( 1+a{{z}^{2}} \right)}}
Integrating both sides we get;
⇒±u+c=12ln⁡((1+az2)−1(1+az2)+1)+(1+az2)\Rightarrow \pm u+c=\dfrac{1}{2}\ln \left( \dfrac{\sqrt{\left( 1+a{{z}^{2}} \right)}-1}{\sqrt{\left( 1+a{{z}^{2}} \right)}+1} \right)+\sqrt{\left( 1+a{{z}^{2}} \right)}
Now, putting value of u;
⇒±(x+ay)+c=12ln⁡((1+az2)−1(1+az2)+1)+(1+az2)\Rightarrow \pm \left( x+ay \right)+c=\dfrac{1}{2}\ln \left( \dfrac{\sqrt{\left( 1+a{{z}^{2}} \right)}-1}{\sqrt{\left( 1+a{{z}^{2}} \right)}+1} \right)+\sqrt{\left( 1+a{{z}^{2}} \right)}

This is the required solution of the partial differential equation.

Note:
Since the given partial differential equation in this question is a non linear partial differential equation, we need to understand it. A non linear partial differential equation is defined as an equation in which the degree of the dependent variable should be other than one and also the degree of the differential equation should not be one.