Question

A Challenging Question :-

${ \bigstar { \underline { \boxed { \red { \bf { \dfrac{d²y}{dt²} + A( t) \dfrac{dy}{dt} + B(t)y(t) = g(t) }}}}}}{\bigstar}$

Find the solution in the form of volterra Integral (:


Standard :- Msc 3rd year​

Answer 1

Answer:

${ \bigstar { \underline { \boxed { \red { \bf { \dfrac{d²y}{dt²} + A( t) \dfrac{dy}{dt} + B(t)y(t) = g(t) }}}}}}{\bigstar}</p><p>$

Multiply t and t to get t2.

$\frac{d(y)}{dt^{2} } dt^{2} \frac{d(y)}{dt} byt = gt$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$\frac{d(y)}{dt^{2} }at^{3} \frac{d(y)}{dt} by = gt$

The equation is in standard form.

$0 = gt$

This is false for any a

$a\:∈ \:$

Answer 2

$\dfrac{d²y}{dt²} + A( t) \dfrac{dy}{dt} + B(t)y(t) = g(t)$

$\dfrac{d²y}{dt²} + At\dfrac{dy}{dt} + Btyt= gt$

${d}^{2-1}y{t}^{-2}+At\times \frac{dy}{dt}+Btyt=gt$

${d}^{1}y{t}^{-2}+At\times \frac{dy}{dt}+Btyt=gt$

$dy{t}^{-2}+At\times \frac{dy}{dt}+Btyt=gt$

$dy\times \frac{1}{{t}^{2}}+At\times \frac{dy}{dt}+Btyt=gt$

$\frac{dy\times 1}{{t}^{2}}+At\times \frac{dy}{dt}+Btyt=gt$

$\frac{dy}{{t}^{2}}+At\times \frac{dy}{dt}+Btyt=gt$

$\frac{dy}{{t}^{2}}+At\times \frac{y}{t}+Btyt=gt$

$\frac{dy}{{t}^{2}}+Ay+Btyt=gt$

$\frac{dy}{{t}^{2}}+Ay+Bt {}^{2} y=gt$

$y(\frac{d}{{t}^{2}}+A+B{t}^{2})=gt$

$\frac{d}{{t}^{2}}+A+B{t}^{2}=\frac{gt}{y}$

$A+B{t}^{2}=\frac{gt}{y}-\frac{d}{{t}^{2}}$

$A=\frac{gt}{y}-\frac{d}{{t}^{2}}-B{t}^{2}$