Differential equations practice · Mixed difficulty

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 B ($y = C{e^{\lambda x}}$) is correct because it represents the solution of the differential equation derived from the given condition about the subnormal.

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.

Option C ($\ln(x+\sqrt{1+x^2})$) is correct because it is the solution to the given differential equation with the initial condition y(0) = 0.

Option B (Homogeneous) is correct because the resulting differential equation can be expressed in the form dy/dx = f(y/x), which is the definition of a homogeneous 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 A (${y^2} + 2xy - {x^2} = c$) is correct because it is the solution obtained by solving the homogeneous differential equation using appropriate substitution and integration.

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 (${y^2} \mp {x^2} = {k^2}$) is correct because it is the solution to the differential equation derived from the given condition about the length of the normal and the radius vector.

Option C ($\sqrt{6}$) is correct because the maximum value of the scalar triple product occurs when $\bar{w}$ is in the direction of $\bar{u} \times \bar{v}$, and its magnitude is $|\bar{u} \times \bar{v}|$.

Option C ($\frac{{1}}{{(x+y)^{2}}}=x^{2}+1+c e^{x^{2}}$) is correct because it's the solution obtained by transforming the given differential equation into a Bernoulli equation and solving it.

Option C (2,2) is correct because the highest order derivative is 2 and its power is 2 after squaring both sides.

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 B ($\frac{{1+3 e^2}}{{2(3 e^2-1)}}$) is correct because it represents the value of $\alpha$ obtained by solving the given differential equation and applying the initial conditions.

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 ($2 y=x(x^2+y^2)$) is correct because it is the solution to the given differential equation with the initial condition y(1)=1.

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 A ($y^2+2 x y-x^2=c$) is correct because it represents the solution to the given differential equation after applying the substitution $y=vx$ and integrating.

Option C ($\frac{\sin y}{x}-e^{\frac{1}{x^2}}=C$) is correct because it is the solution obtained after rearranging and integrating the given differential equation using an integrating factor.

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 B (3) is correct because by using the given functional equation and the differential equation, we find that y = x + 1, and thus y(1/4) + y(3/4) = 3.

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 B ($\sin \left(\frac{y}{x}\right)=\log |c x|$) is correct because it is the solution obtained after solving the given homogeneous differential equation using the substitution $y=vx$.

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.