The equation of the curve satisfying the differential equation $y^{^{''}} (x^{2} + 1) = 4 x y^{^{'}}$ , passing through the point $(0, - 4)$ and having slope of the tangent at $x = 0$ as $4$ is

y = 4 (\frac{x^{5}}{5} + x + \frac{2 x^{3}}{3} - 1)

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

y = 4 (\frac{x^{5}}{5} - x + \frac{2 x^{3}}{3} - 1)

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

y = 4 (\frac{x^{5}}{5} + x + \frac{2 x^{3}}{3} + 1)

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

None of these

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

Open in App

Solution

The correct option is A $y = 4 (\frac{x^{5}}{5} + x + \frac{2 x^{3}}{3} - 1)$
We have, $\frac{y_{2}}{y_{1}} = 2. \frac{2 x}{x^{2} + 1} .$
Integrating with respect to. $x,$ we get
$log y_{1} = 2 log (x^{2} + 1) + log c \Rightarrow y_{1} = c {(x^{2} + 1)}^{2}$
Since $y_{1} (0) = 4 ∴ c = 4$
$∴ y_{1} = 4 (x^{2} + 1) \Rightarrow d y = 4 (x^{4} + 1 + {2 x}^{2}) d x$
Integrating again, we get $y = 5 (\frac{x^{5}}{5} + \frac{{2 x}^{3}}{3} + x - 1) + k$
Putting $x = 0, y = - 4,$ we get $k = - 4.$
$∴$ The required equation of curve is
$y = 4 (\frac{x^{5}}{5} + \frac{{2 x}^{3}}{3} + x - 1) .$

Suggest Corrections

Similar questions

The curve y = ax³ + bx² + cx + 5 touches the x-axis at P(-2, 0) and cuts the y-axis at the point Q where its gradient is 3. the equation of the curve is... .

Q. The equation of the curve satisfying the differential equation

y_{2} (x^{2} + 1) = 2 x y_{1}

passing through the point

(0, 1)

and having slope of tangent at

x = 0

3

Q. A solution curve of the differential equation given by

(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0

passes through (1, 3)

The equation of the tangent to the curve at (1, 3) is

Q. The equation of the curve satisfying the differential equation

y (x + y^{3}) d x = x (y^{3} - x) d y

and passing through the point

(1, 1)

Column-I	Column-II
$(I)$ The curve $C$ which passes through $(1, 1)$ and has differential equation as $(2 x^{2} y - 2 y^{4}) d x + (2 x^{3} + 3 x y^{3}) d y = 0$ is given by $α ln \| x \| + β ln \| y \| + \frac{y^{3}}{x^{2}} = γ$ , then $(α + β + γ)$ equals-	(P) $1$
$(I I)$ A curve $y = f (x)$ passes through $(2, 0)$ and slope of tangent at any point $P (x, y)$ on it equals $\frac{(x + 1)^{2} + (y - 3)}{(x + 1)}$ , then $f (3)$ is-	(Q) $3$
$(I I I)$ Let $f : R^{+} \to R$ satisfies the functional equation $f (x y) = e^{x y - y - x} (e^{y} f (x) + e^{x} f (y))$ for all $x, y \in R^{+}$ . If $f^{'} (1) = e,$ then $ln (ln (f (e)))$ equals-	(R) $5$
$(I V)$ A curve $y = f (x)$ passing through the point $(\frac{1}{2}, \frac{1}{4})$ satisfies the differential equation $\frac{x}{y} \frac{d y}{d x} = \frac{3 x^{2} - y}{2 y - x^{2}}$ is a conic whose length of latus rectum is-	(S) $2$
	(T) $4$

Which of the following is only correct combination?

Join BYJU'S Learning Program

The equation of the curve satisfying the differential equation y′′(x2+1)=4xy′, passing through the point (0,−4) and having slope of the tangent at x=0 as 4 is

The equation of the curve satisfying the differential equation $y^{^{''}} (x^{2} + 1) = 4 x y^{^{'}}$ , passing through the point $(0, - 4)$ and having slope of the tangent at $x = 0$ as $4$ is