Question

y = ax^2+ b x then form a differential equation where a,b are parameters​

Answer 1

Step-by-step explanation:

We have,

$y = a {x}^{2} + bx \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

Differentiating both sides w.r.t x,

$\implies \: {y}_{1} = 2a x + b\: \: \: \: \: \: \: \: \: \: \: \: \: ...(2)$

Again, differentiating both sides w.r.t x,

$\implies \: {y}_{2} = 2a \: \: \: \: \: \: \: \: \: \: \: \: \: ...(3)$

Put (3) in (2),

$\implies \: {y}_{1} =x{y}_{2} + b$

$\implies b = {y}_{1} - x{y}_{2} \: \: \: \: \: \: \: \: \: \: \: \: \: ...(4)$

Put (3),(4) in (1),

$y = \dfrac{{y}_{2} }{2} \cdot{x}^{2} + x \left( {y}_{1} - x {y}_{2}\right)$

$\implies \: 2y = {x}^{2} \cdot{y}_{2} +2 x \left( {y}_{1} - x {y}_{2}\right)$

$\implies \: 2y = {x}^{2} \cdot{y}_{2} +2 x \cdot {y}_{1} -2{x}^{2} \cdot {y}_{2}$

$\implies \: 2y = 2 x \cdot {y}_{1} -{x}^{2} \cdot {y}_{2}$

$\implies \: {x}^{2} \cdot {y}_{2} - 2 x \cdot {y}_{1} + 2y = 0$