The value of the constant $^{'} m^{'}$ and $^{'} c^{'}$ for which $y = m x + c$ is a solution of the differential equation $D^{2} y - 3 D y - 4 y = - 4 x$ .

m = - 1; c = 3 / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m = 1; c = - 3 / 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No such real

m, c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m = 1; c = 3 / 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Is $m = 1; c = - 3 / 4$
We have,
$y = m x + c, ⟹ \frac{d y}{d x} = m, ⟹ \frac{d^{2} y}{d x^{2}} = 0$
Substituting in
$D^{2} y - 3 D y - 4 y = - 4 x$ .
i.e. $\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} - 4 y = - 4 x$
$0 - 3 m - 4 (m x + c) = - 4 x$
$- 3 m - 4 c - 4 m x = - 4 x$
$4 m x + (3 m + 4 c) = 4 x + 0$
Comparing coefiiecient of $x$ we get,
$4 m = 4 ⟹ m = 1$
Comparing constant terms we get
$3 m + 4 c = 0 ⟹ 3 (1) = - 4 c ⟹ c = - 3 / 4$

Suggest Corrections

Similar questions

m

c

y = m x + c

D^{2} y + 3 D y + 4 y = 4 x

(D^{2} y = \frac{d^{2} y}{d x^{2}}, D y = \frac{d y}{d x})

If the equation of the normal is y = mx + c to the parabola $y^{2} = 4 a x$ , then find the value of 'c' in terms of a and m.

n =^{m} C_{2}

^{n} C_{2}

Find the slope (m) and y - intercept (c) of the line $y = 4 x - \frac{1}{2}$ .

For what value of m, the system of equations $m x + 3 y = m - 3, a n d 12 x + m y = m$ will have no solution.

Join BYJU'S Learning Program