Differential Equations: questions with answers

Line 1 is correct. Line 2 is the first wrong line: after expanding, the bracket is a polynomial in $y'$, and the degree is the power of the highest-order derivative $y''$, which is $1$. Order $2$, degree $1$.

Line 1 is correct. Line 2 is the first wrong line: the degree is the power of the HIGHEST-ORDER derivative, not the highest power anywhere. That derivative is of order 2 and appears to the power 1, so the degree is 1.

Line 1 is the first wrong line: $\dfrac{d}{dx}\left[(y')^{ 4 }\right] = 4(y')^{ 3 }y''$, so the equation does involve $y''$. Lines 2 and 3 follow from Line 1. The order is $2$ and the degree (the power of $y''$) is $1$.

Line 1 is a correct observation. Line 2 is the first wrong line: the order is the order of the HIGHEST derivative present, wherever in the equation it is written. Line 3 follows from Line 2. The order is 2.

The condition picks one member of the family of solutions, so it must be applied to the general solution -- the equation containing the arbitrary constant $C$ -- after integrating. Substituting a point into the differential equation itself only gives the slope of the curve at that point and fixes nothing.

Line 1 is the first wrong line: substituting the point into the differential equation gives the slope there, $0$, and says nothing about $C$. The constant is fixed by putting $x = 0$, $y = 2$ into the general solution $-\dfrac{1}{y} = -2x^2 + C$, giving $C = -\dfrac{1}{ 2 }$. Line 2 is a true separation; Line 3 uses the wrong $C$ from Line 1.

Separating, $e^{-y}dy = e^xdx$, so $-e^{-y} = e^x + C_1$ and the general solution is $e^x + e^{-y} = C$. Since $e^x + e^{-y} > 0$ for every $x$ and $y$, only a POSITIVE constant names a real curve: $e^x + e^{-y} = 1$ is a particular solution, while $e^x + e^{-y} = -1$ and $e^x + e^{-y} = 0$ are satisfied by no point at all.

R is true: $e^{-y}dy = e^xdx$ integrates to $-e^{-y} = e^x + C_1$, i.e. $e^x + e^{-y} = C$. But $e^x + e^{-y}$ is positive for every $x, y$, so $C = -2$ is satisfied by no point and $e^x + e^{-y} = -2$ is not a particular solution: A is false.

After separating, the integrals are of powers -- $y^{-2}$, $y^{-\frac{1}{2}}$ -- or of $\dfrac{1}{1 + y^2}$, and only $\dfrac{1}{y}$ integrates to a logarithm: $\int y^{-2}dy = -\dfrac{1}{y}$, $\int y^{-\frac{1}{2}}dy = 2\sqrt y$, $\int\dfrac{dy}{1 + y^2} = \tan^{-1}y$. Hence $\sqrt y - \sqrt x = C$.

Line 1 is correct. Line 2 is the first wrong line: integrating both sides adds an arbitrary constant, $\log|y| = -2\log|x| + \log c$, so the general solution is the family $yx^{ 2 } = c$, not the single curve of Line 3.

$\dfrac{dV}{dt} = 20$ and, by the chain rule, $\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt}$, so $4\pi r^2\dfrac{dr}{dt} = 20$. The radius does NOT change at a constant rate: $\dfrac{dr}{dt} = \dfrac{ 20 }{4\pi r^2}$ falls as the balloon grows.

Line 1 is correct. Line 2 is the first wrong line: the constants $4$ and $5$ have degree $0$ while the other terms have degree $1$, so neither numerator nor denominator is homogeneous and $F(\lambda x, \lambda y) \neq F(x, y)$. Line 3 follows from Line 2.

Lines 1 and 2 are correct. Line 3 is the first wrong line: $v$ was introduced by the substitution and must be replaced by $\dfrac{y}{x}$, giving $y = x\log|x| + Cx$. A solution in $v$ is not a solution of the given equation.

Lines 1 and 2 are correct. Line 3 is the first wrong line: $e^{ 4\log x} = e^{\log x^{ 4 }} = x^{ 4 }$, since $e^{\log f} = f$; it is not $4x$.

Here $x$ is the dependent variable and $y$ the independent one, so the roles in the formula swap: $x\cdot\text{IF} = \int Q\cdot\text{IF}\,dy + C$. It is $x$ that multiplies the factor on the left, and the integration is with respect to $y$.