Find the solution to the given IVP.
\[ -\frac{6}{{y}^{2}} = -2 \, {y} {y'} + \frac{4 \, t}{{y}^{2}} \hspace{2em} y( 3 )= 2 \]
Answer:
\[ {y} = {\left(t^{2} + 3 \, t - 10\right)}^{\frac{1}{3}} \]
Find the solution to the given IVP.
\[ -3 \, {y} \cos\left(t\right) + 3 \, {y'} = 0 \hspace{2em} y( 0 )= -4 \]
Answer:
\[ {y} = -4 \, e^{\sin\left(t\right)} \]
Find the solution to the given IVP.
\[ -\frac{12 \, t}{{y}} - \frac{6}{{y}} = -3 \, {y} {y'} \hspace{2em} y( 2 )= 3 \]
Answer:
\[ {y} = \sqrt{2 \, t^{2} + 2 \, t - 3} \]
Find the solution to the given IVP.
\[ 6 \, {y} = 2 \, {y'} t \hspace{2em} y( -1 )= -1 \]
Answer:
\[ {y} = t^{3} \]
Find the solution to the given IVP.
\[ 0 = -2 \, {y'} t - 6 \, {y} \hspace{2em} y( -2 )= -\frac{1}{4} \]
Answer:
\[ {y} = \frac{2}{t^{3}} \]
Find the solution to the given IVP.
\[ 0 = -2 \, {y'} t + 4 \, {y} \hspace{2em} y( 3 )= 27 \]
Answer:
\[ {y} = 3 \, t^{2} \]
Find the solution to the given IVP.
\[ 2 \, {y} {y'} - \frac{12 \, t}{{y}} - \frac{2}{{y}} = 0 \hspace{2em} y( 2 )= 4 \]
Answer:
\[ {y} = \sqrt{3 \, t^{2} + t + 2} \]
Find the solution to the given IVP.
\[ -4 \, {y} = 2 \, {y'} t \hspace{2em} y( 2 )= \frac{1}{4} \]
Answer:
\[ {y} = \frac{1}{t^{2}} \]
Find the solution to the given IVP.
\[ -2 \, {y'} = -2 \, {y} \cos\left(t\right) \hspace{2em} y( 0 )= -1 \]
Answer:
\[ {y} = -e^{\sin\left(t\right)} \]
Find the solution to the given IVP.
\[ -3 \, {y} \sin\left(t\right) = 3 \, {y'} \hspace{2em} y( 0 )= 4 \, e \]
Answer:
\[ {y} = 4 \, e^{\cos\left(t\right)} \]
Find the solution to the given IVP.
\[ 6 \, {y} {\left(t - 1\right)} = 3 \, {y'} \hspace{2em} y( 0 )= 2 \, e^{\left(-5\right)} \]
Answer:
\[ {y} = 2 \, e^{\left(t^{2} - 2 \, t - 5\right)} \]
Find the solution to the given IVP.
\[ 9 \, {y} = -3 \, {y'} t \hspace{2em} y( -2 )= -\frac{1}{8} \]
Answer:
\[ {y} = \frac{1}{t^{3}} \]
Find the solution to the given IVP.
\[ 3 \, {y'} t + 6 \, {y} = 0 \hspace{2em} y( 2 )= -\frac{1}{2} \]
Answer:
\[ {y} = -\frac{2}{t^{2}} \]
Find the solution to the given IVP.
\[ -4 \, {y} \sin\left(t\right) - 2 \, {y'} = 0 \hspace{2em} y( 0 )= 3 \, e^{2} \]
Answer:
\[ {y} = 3 \, e^{\left(2 \, \cos\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ -2 \, {y'} {y} = -\frac{12 \, t}{{y}^{2}} - \frac{2}{{y}^{2}} \hspace{2em} y( -2 )= 3 \]
Answer:
\[ {y} = {\left(3 \, t^{2} + t + 17\right)}^{\frac{1}{3}} \]
Find the solution to the given IVP.
\[ 3 \, {y'} t = 6 \, {y} \hspace{2em} y( 1 )= -2 \]
Answer:
\[ {y} = -2 \, t^{2} \]
Find the solution to the given IVP.
\[ -2 \, {y'} = 4 \, {y} {\left(t + 1\right)} \hspace{2em} y( 0 )= -e^{2} \]
Answer:
\[ {y} = -e^{\left(-t^{2} - 2 \, t + 2\right)} \]
Find the solution to the given IVP.
\[ 0 = -2 \, {y} {y'} + \frac{12 \, t}{{y}} + \frac{2}{{y}} \hspace{2em} y( -2 )= 4 \]
Answer:
\[ {y} = \sqrt{3 \, t^{2} + t + 6} \]
Find the solution to the given IVP.
\[ -2 \, {y'} t - 4 \, {y} = 0 \hspace{2em} y( -2 )= -\frac{3}{4} \]
Answer:
\[ {y} = -\frac{3}{t^{2}} \]
Find the solution to the given IVP.
\[ 0 = -3 \, {y'} t - 9 \, {y} \hspace{2em} y( 3 )= \frac{1}{27} \]
Answer:
\[ {y} = \frac{1}{t^{3}} \]
Find the solution to the given IVP.
\[ -2 \, {y'} t - 6 \, {y} = 0 \hspace{2em} y( 4 )= -\frac{3}{64} \]
Answer:
\[ {y} = -\frac{3}{t^{3}} \]
Find the solution to the given IVP.
\[ -3 \, {y} \cos\left(t\right) = -3 \, {y'} \hspace{2em} y( 0 )= -3 \]
Answer:
\[ {y} = -3 \, e^{\sin\left(t\right)} \]
Find the solution to the given IVP.
\[ -12 \, {y} t = -3 \, {y'} \hspace{2em} y( 0 )= -4 \]
Answer:
\[ {y} = -4 \, e^{\left(2 \, t^{2}\right)} \]
Find the solution to the given IVP.
\[ 0 = -2 \, {y'} t + 6 \, {y} \hspace{2em} y( 2 )= 32 \]
Answer:
\[ {y} = 4 \, t^{3} \]
Find the solution to the given IVP.
\[ 0 = 3 \, {y'} t + 9 \, {y} \hspace{2em} y( 4 )= \frac{1}{32} \]
Answer:
\[ {y} = \frac{2}{t^{3}} \]
Find the solution to the given IVP.
\[ -6 \, {y} \cos\left(t\right) - 3 \, {y'} = 0 \hspace{2em} y( 0 )= 3 \]
Answer:
\[ {y} = 3 \, e^{\left(-2 \, \sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ \frac{12 \, t}{{y}} + \frac{4}{{y}} = 2 \, {y} {y'} \hspace{2em} y( -3 )= 2 \]
Answer:
\[ {y} = \sqrt{3 \, t^{2} + 2 \, t - 17} \]
Find the solution to the given IVP.
\[ 0 = -6 \, {y} \cos\left(t\right) + 2 \, {y'} \hspace{2em} y( 0 )= -4 \]
Answer:
\[ {y} = -4 \, e^{\left(3 \, \sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 0 = 2 \, {y'} t + 6 \, {y} \hspace{2em} y( -3 )= \frac{4}{27} \]
Answer:
\[ {y} = -\frac{4}{t^{3}} \]
Find the solution to the given IVP.
\[ -3 \, {\left(2 \, t - 1\right)} {y} = -3 \, {y'} \hspace{2em} y( 0 )= 4 \, e^{3} \]
Answer:
\[ {y} = 4 \, e^{\left(t^{2} - t + 3\right)} \]
Find the solution to the given IVP.
\[ -\frac{12 \, t}{{y}} - \frac{6}{{y}} = -2 \, {y}^{2} {y'} \hspace{2em} y( -2 )= 2 \]
Answer:
\[ {y} = {\left(3 \, t^{2} + 3 \, t + 2\right)}^{\frac{1}{3}} \]
Find the solution to the given IVP.
\[ 0 = 6 \, {y} \cos\left(t\right) - 2 \, {y'} \hspace{2em} y( 0 )= -4 \]
Answer:
\[ {y} = -4 \, e^{\left(3 \, \sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 3 \, {y'} = 3 \, {y} {\left(6 \, t + 1\right)} \hspace{2em} y( 0 )= 3 \, e^{\left(-3\right)} \]
Answer:
\[ {y} = 3 \, e^{\left(3 \, t^{2} + t - 3\right)} \]
Find the solution to the given IVP.
\[ -2 \, {y} {\left(6 \, t - 1\right)} = -2 \, {y'} \hspace{2em} y( 0 )= -e^{4} \]
Answer:
\[ {y} = -e^{\left(3 \, t^{2} - t + 4\right)} \]
Find the solution to the given IVP.
\[ 0 = 2 \, {y} \cos\left(t\right) + 2 \, {y'} \hspace{2em} y( 0 )= 2 \]
Answer:
\[ {y} = 2 \, e^{\left(-\sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 0 = 9 \, {y} \sin\left(t\right) + 3 \, {y'} \hspace{2em} y( 0 )= e^{3} \]
Answer:
\[ {y} = e^{\left(3 \, \cos\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 0 = -6 \, {y} \cos\left(t\right) - 2 \, {y'} \hspace{2em} y( 0 )= 2 \]
Answer:
\[ {y} = 2 \, e^{\left(-3 \, \sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 0 = 3 \, {y'} {y} - \frac{12 \, t}{{y}} - \frac{6}{{y}} \hspace{2em} y( 2 )= 2 \]
Answer:
\[ {y} = \sqrt{2 \, t^{2} + 2 \, t - 8} \]
Find the solution to the given IVP.
\[ -3 \, {y'} t + 9 \, {y} = 0 \hspace{2em} y( -2 )= 32 \]
Answer:
\[ {y} = -4 \, t^{3} \]
Find the solution to the given IVP.
\[ 3 \, {y}^{2} {y'} - \frac{9}{{y}} = \frac{12 \, t}{{y}} \hspace{2em} y( -2 )= 3 \]
Answer:
\[ {y} = {\left(2 \, t^{2} + 3 \, t + 25\right)}^{\frac{1}{3}} \]
Find the solution to the given IVP.
\[ -2 \, {\left(2 \, t + 3\right)} {y} + 2 \, {y'} = 0 \hspace{2em} y( 0 )= -4 \, e^{\left(-1\right)} \]
Answer:
\[ {y} = -4 \, e^{\left(t^{2} + 3 \, t - 1\right)} \]
Find the solution to the given IVP.
\[ -2 \, {y'} = -2 \, {y} \sin\left(t\right) \hspace{2em} y( 0 )= -4 \, e^{\left(-1\right)} \]
Answer:
\[ {y} = -4 \, e^{\left(-\cos\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ -9 \, {y} \sin\left(t\right) = -3 \, {y'} \hspace{2em} y( 0 )= -e^{\left(-3\right)} \]
Answer:
\[ {y} = -e^{\left(-3 \, \cos\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ -\frac{4 \, t}{{y}} - \frac{2}{{y}} = -2 \, {y} {y'} \hspace{2em} y( 2 )= 3 \]
Answer:
\[ {y} = \sqrt{t^{2} + t + 3} \]
Find the solution to the given IVP.
\[ -3 \, {y'} t + 9 \, {y} = 0 \hspace{2em} y( 4 )= -64 \]
Answer:
\[ {y} = -t^{3} \]
Find the solution to the given IVP.
\[ 0 = 3 \, {y'} t + 6 \, {y} \hspace{2em} y( 4 )= \frac{1}{8} \]
Answer:
\[ {y} = \frac{2}{t^{2}} \]
Find the solution to the given IVP.
\[ -6 \, {y} = 3 \, {y'} t \hspace{2em} y( 1 )= 4 \]
Answer:
\[ {y} = \frac{4}{t^{2}} \]
Find the solution to the given IVP.
\[ 6 \, {y} \cos\left(t\right) + 2 \, {y'} = 0 \hspace{2em} y( 0 )= 1 \]
Answer:
\[ {y} = e^{\left(-3 \, \sin\left(t\right)\right)} \]
Find the solution to the given IVP.
\[ 0 = 2 \, t {y'} + 4 \, {y} \hspace{2em} y( 1 )= 2 \]
Answer:
\[ {y} = \frac{2}{t^{2}} \]
Find the solution to the given IVP.
\[ 0 = 3 \, {y'} t - 9 \, {y} \hspace{2em} y( 2 )= 32 \]
Answer:
\[ {y} = 4 \, t^{3} \]