With initial condition x(1) = 0.5 , the solution of the differential equation \(t\frac{dx}{dt}+x=t\) is
\(x=\frac{t}{2}\)
The left side is d(tx)/dt, so tx=t²/2+C. The condition x(1)=1/2 gives C=0. Hence x=t/2 on the interval containing t=1.
Important Questions from Differential Equations
The Green's function for the differential equation \(\rm\frac{d^2x}{dt^2}\) + x = f(t), satisfying the initial conditions x(0) = \(\rm\frac{dx}{dt}\) (0) = 0, is
G(t, τ) = \(\begin{cases}0 & \text { for } \quad 0<\rm t<\tau \\ \sin (\rm t−\tau) & \text { for } \quad \rm t>\tau\end{cases}\)
The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
- \(\smallint \frac{{dx}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}}\) = ?
A differential equation is given as x (t + 2) + 3x (t + 1) + 2x (t) =0; x(0) = 0, x(1) = 1. The solution of this equation will be:
If y = \(\rm\left(\frac{1}{x}\right)^x \), then value of \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\) is:
If \(f\left( x \right) = \frac{{ax + b}}{{cx + d}}\) and f(f(x)) = x then
