Solveeit Logo
Login

Question

Question: Write the order and degree of the differential equation \[y = x\dfrac{{dy}}{{dx}} + a\sqrt {1 + {{\l...

Write the order and degree of the differential equation y=xdydx+a1+(dydx)2y = x\dfrac{{dy}}{{dx}} + a\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}

Explanation

Solution

We use the concept of order and degree of a differential equation and apply them on the given equations to find the answer. Shift all the values except square root one side of the equation and use the formula of (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2ab to solve the required side.

Complete step-by-step answer:
We have to find the order and degree of the differential equationy=xdydx+a1+(dydx)2y = x\dfrac{{dy}}{{dx}} + a\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} ……………….... (1)
Shift the termxdydxx\dfrac{{dy}}{{dx}} from RHS to LHS of the equation
⇒y−xdydx=a1+(dydx)2\Rightarrow y - x\dfrac{{dy}}{{dx}} = a\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}
Square both sides of the equation
⇒(y−xdydx)2=(a1+(dydx)2)2\Rightarrow {\left( {y - x\dfrac{{dy}}{{dx}}} \right)^2} = {\left( {a\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } \right)^2}
Use the formula (a−b)2=a2+b2−2ab{(a - b)^2} = {a^2} + {b^2} - 2abto expand LHS
⇒(y)2+(xdydx)2−2(y)(xdydx)=a2(1+(dydx)2)2\Rightarrow {\left( y \right)^2} + {\left( {x\dfrac{{dy}}{{dx}}} \right)^2} - 2\left( y \right)\left( {x\dfrac{{dy}}{{dx}}} \right) = {a^2}{\left( {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } \right)^2}
⇒y2+x2(dydx)2−2xy(dydx)=a2(1+(dydx)2)2\Rightarrow {y^2} + {x^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) = {a^2}{\left( {\sqrt {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} } \right)^2}
Cancel square root by square power on RHS of the equation.
⇒y2+x2(dydx)2−2xy(dydx)=a2(1+(dydx)2)\Rightarrow {y^2} + {x^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) = {a^2}\left( {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right)
⇒y2+x2(dydx)2−2xy(dydx)=a2+a2(dydx)2\Rightarrow {y^2} + {x^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) = {a^2} + {a^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2}
Shift all values to LHS of the equation
⇒y2+x2(dydx)2−2xy(dydx)−a2−a2(dydx)2=0\Rightarrow {y^2} + {x^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) - {a^2} - {a^2}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 0
Collect terms with same variables
\Rightarrow \left\\{ {{x^2}{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2} - {a^2}{{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right\\} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) + {y^2} - {a^2} = 0
\Rightarrow \left\\{ {{x^2} - {a^2}} \right\\}{\left( {\dfrac{{dy}}{{dx}}} \right)^2} - 2xy\left( {\dfrac{{dy}}{{dx}}} \right) + {y^2} - {a^2} = 0...................… (2)
Now we use the definition of order and degree on equation (2)
Since, highest order derivative is dydx\dfrac{{dy}}{{dx}}
⇒\RightarrowOrder =1 = 1
Also, the highest power of the highest order derivative is given by (dydx)2{\left( {\dfrac{{dy}}{{dx}}} \right)^2}
⇒\RightarrowDegree =2 = 2

∴\therefore Order of the differential equation is 1 and the degree of the differential equation is 2.

Note: * Order of a differential equation is the order of the highest derivative. Example: In a differential equation d3ydx3−3d2ydx2+dydx−2y=3\dfrac{{{d^3}y}}{{d{x^3}}} - 3\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} - 2y = 3has order 3, as the highest derivative of third order.

  • Degree of a differential equation is the highest power of the highest order derivative. Example: In a differential equation(d3ydx3)2−3d2ydx2+(dydx)4−2y=3{\left( {\dfrac{{{d^3}y}}{{d{x^3}}}} \right)^2} - 3\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^4} - 2y = 3 has order 3 and degree 2, as the power of the highest order is 2.