Solving Separable DEs with Initial Conditions

Use separation of variables to find the particular solution of a first-order ODE satisfying a given initial condition, typically as the main task or a major part of a multi-part problem.

jee-main 2023 Q76 View
The area enclosed by the closed curve $C$ given by the differential equation $\frac{dy}{dx} + \frac{x + a}{y - 2} = 0$, $y(1) = 0$ is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is
(1) $2\sqrt{3}$
(2) $\frac{2\sqrt{3}}{3}$
(3) 2
(4) $\frac{4\sqrt{3}}{3}$
jee-main 2023 Q84 View
Let $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ be the solution of the differential equation $y ( x + 1 ) d x - x ^ { 2 } d y = 0 , y ( 1 ) = e$. Then $\lim _ { x \rightarrow 0 ^ { + } } f ( x )$ is equal to
(1) 0
(2) $\frac { 1 } { e }$
(3) $e ^ { 2 }$
(4) $\frac { 1 } { e ^ { 2 } }$
jee-main 2023 Q84 View
The solution of the differential equation $\frac{dy}{dx} = -\left(\frac{x^{2} + 3y^{2}}{3x^{2} + y^{2}}\right)$, $y(1) = 0$ is
(1) $\log_{e}|x + y| - \frac{xy}{(x+y)^{2}} = 0$
(2) $\log_{e}|x + y| + \frac{xy}{(x+y)^{2}} = 0$
(3) $\log_{e}|x + y| + \frac{2xy}{(x+y)^{2}} = 0$
(4) $\log_{e}|x + y| - \frac{2xy}{(x+y)^{2}} = 0$
jee-main 2023 Q84 View
If the solution curve of the differential equation $\left( y - 2 \log _ { e } x \right) d x + \left( x \log _ { e } x ^ { 2 } \right) d y = 0 , x > 1$ passes through the points $\left( e , \frac { 4 } { 3 } \right)$ and $\left( e ^ { 4 } , \alpha \right)$, then $\alpha$ is equal to $\_\_\_\_$
jee-main 2023 Q74 View
The slope of tangent at any point $(x, y)$ on a curve $y = y(x)$ is $\frac{x^2 + y^2}{2xy}$, $x > 0$. If $y(2) = 0$, then a value of $y(8)$ is
(1) $-4\sqrt{2}$
(2) $2\sqrt{3}$
(3) $-2\sqrt{3}$
(4) $4\sqrt{3}$
jee-main 2023 Q83 View
Let $y = y ( x ) , y > 0$, be a solution curve of the differential equation $\left( 1 + x ^ { 2 } \right) d y = y ( x - y ) d x$. If $y ( 0 ) = 1$ and $y ( 2 \sqrt { 2 } ) = \beta$, then
(1) $e ^ { 3 \beta - 1 } = e ( 3 + 2 \sqrt { 2 } )$
(2) $e ^ { 3 \beta - 1 } = e ( 5 + \sqrt { 2 } )$
(3) $e ^ { \beta - 1 } = e ^ { - 2 } ( 3 + 2 \sqrt { 2 } )$
(4) $e ^ { \beta - 1 } = e ^ { - 2 } ( 5 + \sqrt { 2 } )$
jee-main 2024 Q76 View
Let $\alpha$ be a non-zero real number. Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $f(0) = 1$ and $\lim_{x \to -\infty} f(x) = 1$. If $f'(x) = \alpha f(x) + 3$, for all $x \in \mathbb{R}$, then $f(-\log_e 2)$ is equal to:
(1) 1
(2) 5
(3) 9
(4) 7
jee-main 2024 Q76 View
If $\sin \left( \frac { y } { x } \right) = \log _ { e } | x | + \frac { \alpha } { 2 }$ is the solution of the differential equation $x \cos \left( \frac { y } { x } \right) \frac { d y } { d x } = y \cos \left( \frac { y } { x } \right) + x$ and $y ( 1 ) = \frac { \pi } { 3 }$, then $\alpha ^ { 2 }$ is equal to
(1) 3
(2) 12
(3) 4
(4) 9
jee-main 2024 Q75 View
The solution curve of the differential equation $y \frac { d x } { d y } = x \left( \log _ { e } x - \log _ { e } y + 1 \right) , \quad x > 0 , \quad y > 0$ passing through the point $(e, 1)$ is
(1) $\log _ { e } \frac { y } { x } = x$
(2) $\log _ { e } \frac { y } { x } = y ^ { 2 }$
(3) $\log _ { e } \frac { x } { y } = y$
(4) $2 \log _ { e } \frac { x } { y } = y + 1$
jee-main 2024 Q88 View
If the solution of the differential equation $( 2 x + 3 y - 2 ) d x + ( 4 x + 6 y - 7 ) d y = 0 , y ( 0 ) = 3$, is $\alpha x + \beta y + 3 \log _ { e } | 2 x + 3 y - \gamma | = 6$, then $\alpha + 2 \beta + 3 \gamma$ is equal to $\_\_\_\_$.
jee-main 2024 Q76 View
If $y = y ( x )$ is the solution curve of the differential equation $x ^ { 2 } - 4 d y - y ^ { 2 } - 3 y d x = 0 , x > 2 , y ( 4 ) = \frac { 3 } { 2 }$ and the slope of the curve is never zero, then the value of $\mathrm { y } ( 10 )$ equals :
(1) $\frac { 3 } { 1 + ( 8 ) ^ { \frac { 1 } { 4 } } }$
(2) $\frac { 3 } { 1 + 2 \sqrt { z } }$
(3) $\frac { 3 } { 1 - 2 \sqrt { 2 } }$
(4) $\frac { 3 } { 1 - ( 8 ) ^ { \frac { 1 } { 4 } } }$
jee-main 2024 Q89 View
If the solution curve, of the differential equation $\frac { d y } { d x } = \frac { x + y - 2 } { x - y }$ passing through the point $( 2,1 )$ is $\tan ^ { - 1 } \frac { y - 1 } { x - 1 } - \frac { 1 } { \beta } \log _ { e } \alpha + \frac { y - 1 } { x - 1 } ^ { 2 } = \log _ { e } x - 1$, then $5 \beta + \alpha$ is equal to $\_\_\_\_$.
jee-main 2024 Q88 View
Let $y = y ( x )$ be the solution of the differential equation $( x + y + 2 ) ^ { 2 } d x = d y , y ( 0 ) = - 2$. Let the maximum and minimum values of the function $y = y ( x )$ in $\left[ 0 , \frac { \pi } { 3 } \right]$ be $\alpha$ and $\beta$, respectively. If $( 3 \alpha + \pi ) ^ { 2 } + \beta ^ { 2 } = \gamma + \delta \sqrt { 3 } , \gamma , \delta \in \mathbb { Z }$, then $\gamma + \delta$ equals $\_\_\_\_$
jee-main 2024 Q76 View
Suppose the solution of the differential equation $\frac { d y } { d x } = \frac { ( 2 + \alpha ) x - \beta y + 2 } { \beta x - 2 \alpha y - ( \beta \gamma - 4 \alpha ) }$ represents a circle passing through origin. Then the radius of this circle is:
(1) 2
(2) $\sqrt { 17 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { \sqrt { 17 } } { 2 }$
jee-main 2024 Q89 View
Let $\alpha | x | = | y | \mathrm { e } ^ { x y - \beta } , \alpha , \beta \in \mathbf { N }$ be the solution of the differential equation $x \mathrm {~d} y - y \mathrm {~d} x + x y ( x \mathrm {~d} y + y \mathrm {~d} x ) = 0$, $y ( 1 ) = 2$. Then $\alpha + \beta$ is equal to $\_\_\_\_$
jee-main 2025 Q12 View
Let $x = x ( y )$ be the solution of the differential equation $y = \left( x - y \frac { \mathrm {~d} x } { \mathrm {~d} y } \right) \sin \left( \frac { x } { y } \right) , y > 0$ and $x ( 1 ) = \frac { \pi } { 2 }$. Then $\cos ( x ( 2 ) )$ is equal to :
(1) $1 - 2 \left( \log _ { e } 2 \right) ^ { 2 }$
(2) $1 - 2 \left( \log _ { \mathrm { e } } 2 \right)$
(3) $2 \left( \log _ { e } 2 \right) - 1$
(4) $2 \left( \log _ { e } 2 \right) ^ { 2 } - 1$
jee-main 2025 Q17 View
Let $y = y(x)$ be the solution of the differential equation $\left(xy - 5x^2\sqrt{1+x^2}\right)dx + \left(1+x^2\right)dy = 0$, $y(0) = 0$. Then $y(\sqrt{3})$ is equal to
(1) $\sqrt{\frac{15}{2}}$
(2) $\frac{5\sqrt{3}}{2}$
(3) $2\sqrt{2}$
(4) $\sqrt{\frac{14}{3}}$
jee-main 2025 Q6 View
Let a curve $y = f ( x )$ pass through the points $( 0,5 )$ and $\left( \log _ { e } 2 , k \right)$. If the curve satisfies the differential equation $2 ( 3 + y ) e ^ { 2 x } d x - \left( 7 + e ^ { 2 x } \right) d y = 0$, then $k$ is equal to
(1) 4
(2) 32
(3) 8
(4) 16
jee-main 2025 Q7 View
Let $f : (0, \infty) \rightarrow \mathbf{R}$ be a function which is differentiable at all points of its domain and satisfies the condition $x^{2} f'(x) = 2x f(x) + 3$, with $f(1) = 4$. Then $2f(2)$ is equal to:
(1) 39
(2) 19
(3) 29
(4) 23
jee-main 2025 Q77 View
Q77. If the solution $y = y ( x )$ of the differential equation $\left( x ^ { 4 } + 2 x ^ { 3 } + 3 x ^ { 2 } + 2 x + 2 \right) \mathrm { d } y - \left( 2 x ^ { 2 } + 2 x + 3 \right) \mathrm { d } x = 0$ satisfies $y ( - 1 ) = - \frac { \pi } { 4 }$, then $y ( 0 )$ is equal to :
(1) $\frac { \pi } { 2 }$
(2) $- \frac { \pi } { 2 }$
(3) 0
(4) $\frac { \pi } { 4 }$
jee-main 2025 Q88 View
Q88. Let $y = y ( x )$ be the solution of the differential equation $( x + y + 2 ) ^ { 2 } d x = d y , y ( 0 ) = - 2$. Let the maximum and minimum values of the function $y = y ( x )$ in $\left[ 0 , \frac { \pi } { 3 } \right]$ be $\alpha$ and $\beta$, respectively. If $( 3 \alpha + \pi ) ^ { 2 } + \beta ^ { 2 } = \gamma + \delta \sqrt { 3 } , \gamma , \delta \in \mathbb { Z }$, then $\gamma + \delta$ equals $\_\_\_\_$
jee-main 2025 Q89 View
Q89. Let $\alpha | x | = | y | \mathrm { e } ^ { x y - \beta } , \alpha , \beta \in \mathbf { N }$ be the solution of the differential equation $x \mathrm {~d} y - y \mathrm {~d} x + x y ( x \mathrm {~d} y + y \mathrm {~d} x ) = 0$, $y ( 1 ) = 2$. Then $\alpha + \beta$ is equal to $\_\_\_\_$
jee-main 2025 Q75 View
Q75. The solution curve, of the differential equation $2 y \frac { \mathrm {~d} y } { \mathrm {~d} x } + 3 = 5 \frac { \mathrm {~d} y } { \mathrm {~d} x }$, passing through the point $( 0,1 )$ is a conic, whose vertex lies on the line:
(1) $2 x + 3 y = 9$
(2) $2 x + 3 y = - 9$
(3) $2 x + 3 y = - 6$
(4) $2 x + 3 y = 6$
jee-main 2025 Q76 View
Q76. The solution of the differential equation $\left( x ^ { 2 } + y ^ { 2 } \right) \mathrm { d } x - 5 x y \mathrm {~d} y = 0 , y ( 1 ) = 0$, is :
(1) $\left| x ^ { 2 } - 2 y ^ { 2 } \right| ^ { 6 } = x$
(2) $\left| x ^ { 2 } - 4 y ^ { 2 } \right| ^ { 6 } = x$
(3) $\left| x ^ { 2 } - 4 y ^ { 2 } \right| ^ { 5 } = x ^ { 2 }$
(4) $\left| x ^ { 2 } - 2 y ^ { 2 } \right| ^ { 5 } = x ^ { 2 }$
tmua 2023 Q19 1 marks View
The solution to the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = | - 6 x | \quad \text { for all } x$$
is $y = f ( x ) + c$, where $c$ is a constant.
Which one of the following is a correct expression for $f ( x )$ ?