Differential equations practice · Easy

Option A (3) is correct because the general equation of a parabola with directrix parallel to the x-axis has 3 independent constants, thus the order of the differential equation is 3.

Option C ((-1/2, 4, 0)) is correct because it satisfies the given conditions: b = 2a and a is perpendicular to c.

Option B ($\sin x + \cos y = c$) is correct because it is the solution obtained by separating variables and integrating the given differential equation.

Option B ($y+\sqrt{x^2+y^2}=c x^2$) is the correct solution to the homogeneous differential equation, solved by substituting $y=vx$.

Option B ($\log \tan \frac{y}{2}+2 \sin x=\mathrm{C}$) is correct because it is the solution obtained by using trigonometric identities and integrating the differential equation.

Option A ($x=X+1, y=Y+0$) is correct because it is the substitution that reduces the given differential equation to a homogeneous form.

Option C ($\sqrt {{{(1 - {x^2})}}}$) is correct because it is the integrating factor derived from the given differential equation after rewriting it in standard form.

Option B ($y=\frac{1}{18}(3 x^3+2 e^{3 x}-2)$) is correct because it satisfies the given differential equation and the initial conditions $y_1(0)=\frac{1}{3}$ and $y(0)=0$.

Option C (2,1) is incorrect. Option D (1,2) is incorrect. Option B (1,1) is correct because the order and degree of the differential equation are both 1 after simplification.

Option C ($-\frac{1}{x y}+\log \left(\frac{y}{x}\right)=C$) is correct because it is the solution obtained by integrating the given differential equation after rearrangement.

Option C (Second order and first degree) is correct because the differential equation obtained after eliminating the arbitrary constants is of second order and first degree.

Option D ($(a+m) y=e^{m x}+c e^{-a x}(a+m)$) is correct because it represents the general solution to the given linear differential equation.

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the given differential equation matches the form described in the reason, where P and Q are functions of x or constants.

Option D ($4 \sqrt{3}$) is correct because it is the solution to the differential equation given the initial condition y(2) = 0.

Option C (e/4) is correct because solving the linear differential equation and applying the given condition yields y(e) = e/4.

Option B (a circle of radius one) is correct because the solution to the differential equation, after applying the given condition, results in the equation of a circle with radius one.

Option A (a linear equation) is correct because the resulting differential equation is a linear equation in terms of dx/dy.

Option D is correct because the degree of a differential equation is not defined if the equation is not a polynomial in all differential coefficients, which is the case here due to the logarithm.

Option A ($\left( {x + 1} \right)\frac{{dy}}{{dx}} - y = 0$) is correct because it represents the differential equation of a family of lines where the slope is equal to the y-intercept.

Option B (10) is the correct value of b because it satisfies the given differential equation, the point (1,2), and the integral condition.

Option C ($y\frac{{{d^2}y}}{{d{x^2}}} + {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0$) is correct because it is the differential equation obtained by differentiating the given equation twice and eliminating the arbitrary constant 'a'.

Option D, a hyperbola, is correct. The condition that the tangent is bisected at the point of contact leads to the differential equation xy = C, which represents a hyperbola.

Option D is correct because the degree of a differential equation is not defined if the equation is not a polynomial in all differential coefficients, which is the case here due to the sine function.

Option C (A rectangular hyperbola) is correct because the given condition leads to the differential equation $\frac{{dy}}{{dx}} = \frac{x}{y}$, which integrates to the equation of a rectangular hyperbola.

The correct answer is C ($y=C e^{(-1 / 2) x^2}-4$) because it is the general solution obtained by solving the linear differential equation using an integrating factor.

Option B ($y=x \frac{{d y}}{{d x}}+\frac{{d y}}{{d x}}-\left(\frac{{d y}}{{d x}}\right)^3$) is correct because substituting $\frac{{dy}}{{dx}} = c$ into the original equation yields this result.

Option D ($4{\tan ^{ - 1}}\left( {\frac{{2y + 1}}{{\sqrt 3 }}} \right) = \sqrt 3 \left( {2x + {x^2}} \right) + c$) is correct because it is the solution obtained by separating variables and integrating the given differential equation.

Option C ($y\frac{{{d^2}y}}{{d{x^2}}} - {\left( {\frac{{dy}}{{dx}}} \right)^2} = 0$) is correct because it represents the differential equation after eliminating the arbitrary constants from the given equation.

Option B ($y_{1}+y_{1}^{3}-x y_{2}=0$) is correct because it represents the differential equation of all circles with centers on the y-axis.

Option A (${y^2} = {x^2} - 2x + 2$) is correct because it is the solution to the given differential equation that satisfies the initial condition y(1) = 1.