The ODE to solve is \begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +4\,y \left ( x \right ) =20 \end{align*}
with initial conditions \(y \left ( 0 \right ) =2\).
Trying separable ODE.
In canonical form, the ODE is written as \begin{align*} y' &= F(x,y)\\ &= -4\,y+20 \end{align*}
The ODE \( \frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = -4\,y+20\), is separable. It can be written as \[ \frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = f( x) g(y) \] Where \(f(x)=1\) and \(g(y)=-4\,y+20\). Therefore
\[ \frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = -4\,y+20 \] Hence \begin{align*} \left ( -4\,y+20 \right ) ^{-1}\mathop{\mathrm{d}y}&= \mathop{\mathrm{d}x}\\ \int \left ( -4\,y+20 \right ) ^{-1}\mathop{\mathrm{d}y}&= \int \mathop{\mathrm{d}x}\\ -1/2\,\ln \left ( 2 \right ) -1/4\,\ln \left ( \left | y-5 \right | \right ) &=x+C_{{1}}\\ \end{align*}
Solving for \(y\) gives \[ y = -1/4\,{{\rm e}^{-4\,x-4\,C_{{1}}}}+5 \] The solution above can be written as \begin{equation} \setlength{\fboxsep }{2.5\fboxsep }\boxed{y = -1/4\,C_{{1}}{{\rm e}^{-4\,x}}+5} \end{equation}
Initial conditions are now used to solve for \(C_{{1}}\). Substituting \(x=0\) and \(y=2\) in the above solution gives an equation to solve for the constant of integration. \begin{align*} 2&= -1/4\,C_{{1}}{{\rm e}^{0}}+5\\ &=-1/4\,C_{{1}}+5 \end{align*}
Hence \[ C_{{1}} = 12\, \left ({{\rm e}^{0}} \right ) ^{-1} \] Which is simplified to \[ C_{{1}} = 12 \] Substituting \(C_1\) found above back in the solution gives \[ y \left ( x \right ) = -3\,{{\rm e}^{-4\,x}}+5 \]
The ODE to solve is \begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +3\,y \left ( x \right ) =\sin \left ( x \right ) \end{align*}
with initial conditions \(y \left ( \pi /2 \right ) =3/10\).
Trying Linear ODE.
In canonical form, the ODE is written as \begin{align*} y' &= F(x,y)\\ &= -3\,y+\sin \left ( x \right ) \end{align*}
The ODE is linear in \(y\) and has the form \[ y' = y f(x) + g(x) \] Where \(f(x) = -3\) and \(g(x) = \sin \left ( x \right ) \).
Writing the ODE as \begin{align*} y' - \left (-3\,y\right ) &= \sin \left ( x \right ) \\ y' + 3\,y &= \sin \left ( x \right ) \end{align*}
Therefore the integrating factor \(\mu \) is \[ \mu = e^{\int 3\mathop{\mathrm{d}x}} ={{\rm e}^{3\,x}} \]The ode becomes \begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu y &= \mu \left (\sin \left ( x \right ) \right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (y{{\rm e}^{3\,x}}\right ) &= \sin \left ( x \right ){{\rm e}^{3\,x}}\\ \mathrm{d} \left (y{{\rm e}^{3\,x}}\right ) &= \left (\sin \left ( x \right ){{\rm e}^{3\,x}}\right ) \mathrm{d} x \end{align*}
Integrating both sides gives \begin{align*} y{{\rm e}^{3\,x}} &= -1/10\,\cos \left ( x \right ){{\rm e}^{3\,x}}+3/10\,\sin \left ( x \right ){{\rm e}^{3\,x}} + C_1 \end{align*}
Dividing both sides by the integrating factor \(\mu ={{\rm e}^{3\,x}}\) results in \[ y ={\frac{-1/10\,\cos \left ( x \right ){{\rm e}^{3\,x}}+3/10\,\sin \left ( x \right ){{\rm e}^{3\,x}}}{{{\rm e}^{3\,x}}}}+{\frac{C_{{1}}}{{{\rm e}^{3\,x}}}} \]Simplifying the solution gives \[ \setlength{\fboxsep }{2.5\fboxsep }\boxed{y=3/10\,\sin \left ( x \right ) -1/10\,\cos \left ( x \right ) +C_{{1}}{{\rm e}^{-3\,x}}} \]
Initial conditions are now used to solve for \(C_{{1}}\). Substituting \(x=\pi /2\) and \(y=3/10\) in the above solution gives an equation to solve for the constant of integration. \begin{align*} 3/10&= 3/10\,\sin \left ( \pi /2 \right ) -1/10\,\cos \left ( \pi /2 \right ) +C_{{1}}{{\rm e}^{-3/2\,\pi }}\\ &=3/10+C_{{1}}{{\rm e}^{-3/2\,\pi }} \end{align*}
Hence \[ C_{{1}} = -1/10\,{\frac{3\,\sin \left ( \pi /2 \right ) -\cos \left ( \pi /2 \right ) -3}{{{\rm e}^{-3/2\,\pi }}}} \] Which is simplified to \[ C_{{1}} = 0 \] Substituting \(C_1\) found above back in the solution gives \[ y \left ( x \right ) = 3/10\,\sin \left ( x \right ) -1/10\,\cos \left ( x \right ) \]
The ODE to solve is \begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) -y \left ( x \right ) \left ( 1+3\,{x}^{-1} \right ) =x+2 \end{align*}
with initial conditions \(y \left ( 1 \right ) ={\rm e}-1\).
Trying Linear ODE.
In canonical form, the ODE is written as \begin{align*} y' &= F(x,y)\\ &={\frac{{x}^{2}+xy+2\,x+3\,y}{x}} \end{align*}
The ODE is linear in \(y\) and has the form \[ y' = y f(x) + g(x) \] Where \(f(x) ={\frac{x+3}{x}}\) and \(g(x) ={\frac{{x}^{2}+2\,x}{x}}\).
Writing the ODE as \begin{align*} y' - \left ({\frac{ \left ( x+3 \right ) y}{x}}\right ) &={\frac{{x}^{2}+2\,x}{x}}\\ y' -{\frac{ \left ( x+3 \right ) y}{x}} &={\frac{{x}^{2}+2\,x}{x}} \end{align*}
Therefore the integrating factor \(\mu \) is \[ \mu = e^{\int -{\frac{x+3}{x}}\mathop{\mathrm{d}x}} ={{\rm e}^{-x-3\,\ln \left ( x \right ) }} \]The ode becomes \begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu y &= \mu \left ({\frac{{x}^{2}+2\,x}{x}}\right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (y{{\rm e}^{-x-3\,\ln \left ( x \right ) }}\right ) &={\frac{ \left ({x}^{2}+2\,x \right ){{\rm e}^{-x-3\,\ln \left ( x \right ) }}}{x}}\\ \mathrm{d} \left (y{{\rm e}^{-x-3\,\ln \left ( x \right ) }}\right ) &= \left ({\frac{ \left ({x}^{2}+2\,x \right ){{\rm e}^{-x-3\,\ln \left ( x \right ) }}}{x}}\right ) \mathrm{d} x \end{align*}
Integrating both sides gives \begin{align*} y{{\rm e}^{-x-3\,\ln \left ( x \right ) }} &= -{{\rm e}^{-x-3\,\ln \left ( x \right ) }}x + C_1 \end{align*}
Dividing both sides by the integrating factor \(\mu ={{\rm e}^{-x-3\,\ln \left ( x \right ) }}\) results in \[ y = -x+{\frac{C_{{1}}}{{{\rm e}^{-x-3\,\ln \left ( x \right ) }}}} \]Simplifying the solution gives \[ \setlength{\fboxsep }{2.5\fboxsep }\boxed{y=-x+C_{{1}}{x}^{3}{{\rm e}^{x}}} \]
Initial conditions are now used to solve for \(C_{{1}}\). Substituting \(x=1\) and \(y={\rm e}-1\) in the above solution gives an equation to solve for the constant of integration. \begin{align*}{\rm e}-1&=-1+C_{{1}}{\rm e} \end{align*}
Hence \[ C_{{1}} = 1 \] Substituting \(C_1\) found above back in the solution gives \[ y \left ( x \right ) = -x+{x}^{3}{{\rm e}^{x}} \]
The ODE to solve is \begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +1/3\,y \left ( x \right ) =1/3\, \left ( 1-2\,x \right ) \left ( y \left ( x \right ) \right ) ^{4} \end{align*}
Trying Bernoulli ODE.
In canonical form, the ODE is written as \begin{align*} y' &= F(x,y)\\ &= -y/3-2/3\,{y}^{4}x+1/3\,{y}^{4} \end{align*}
This is a Bernoulli ODE. Comparing the ODE to solve \[ y' = -y/3-2/3\,{y}^{4}x+1/3\,{y}^{4} \] With Bernoulli ODE standard form \[ y' = f_0(x)y+f_1(x)y^n \]
Shows that \(f_0(x)=-1/3\) and \(f_1(x)=-2/3\,x+1/3\) and \(n=4\).
Dividing the ODE by \({y}^{4}\) gives \begin{align*} y'{y}^{-4} &= -1/3{y}^{-3} + -2/3\,x+1/3 \tag{1} \end{align*}
Let \begin{align*} v &={y}^{-3} \tag{2} \end{align*}
Taking derivative of (2) w.r.t \(x\) gives \begin{align*} v' &= -3\,{y}^{-4}y' \\{y}^{-4}&= \frac{v'}{-3\,y'} \tag{3} \end{align*}
Substituting (3) into (1) gives \begin{align*} \frac{v'}{(-3)} &= \left (-1/3\right ) v + -2/3\,x+1/3\\ v' &= (-3)\left (-1/3\right ) v + (-3)\left (-2/3\,x+1/3\right ) \\ &= v+2\,x-1 \end{align*}
The above now is a linear ODE in \(v(x)\) which can be easily solved using an integrating factor.
In canonical form, the ODE is written as \begin{align*} v' &= F(x,v)\\ &= v+2\,x-1 \end{align*}
The ODE is linear in \(v\) and has the form \[ v' = v f(x) + g(x) \] Where \(f(x) = 1\) and \(g(x) = 2\,x-1\).
Writing the ODE as \begin{align*} v' - \left (v\right ) &= 2\,x-1\\ v' -v &= 2\,x-1 \end{align*}
Therefore the integrating factor \(\mu \) is \[ \mu = e^{\int -1\mathop{\mathrm{d}x}} ={{\rm e}^{-x}} \]The ode becomes \begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu v &= \mu \left (2\,x-1\right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (v{{\rm e}^{-x}}\right ) &= \left ( 2\,x-1 \right ){{\rm e}^{-x}}\\ \mathrm{d} \left (v{{\rm e}^{-x}}\right ) &= \left ( \left ( 2\,x-1 \right ){{\rm e}^{-x}}\right ) \mathrm{d} x \end{align*}
Integrating both sides gives \begin{align*} v{{\rm e}^{-x}} &= - \left ( 2\,x+1 \right ){{\rm e}^{-x}} + C_1 \end{align*}
Dividing both sides by the integrating factor \(\mu ={{\rm e}^{-x}}\) results in \[ v = -2\,x-1+{\frac{C_{{1}}}{{{\rm e}^{-x}}}} \]Simplifying the solution gives \[ \setlength{\fboxsep }{2.5\fboxsep }\boxed{v=-2\,x-1+C_{{1}}{{\rm e}^{x}}} \]
Replacing \(v\) in the above by \({y}^{-3}\) from equation (2), gives the final solution.
\[{y}^{-3}=-2\,x-1+C_{{1}}{{\rm e}^{x}} \] Solving for \(y\) gives \begin{align*} y &={\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ y &=-1/2\,{\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}+{\frac{i/2\sqrt{3}}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ y &=-1/2\,{\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}-{\frac{i/2\sqrt{3}}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ \end{align*}
The ODE to solve is \begin{align*} m{\frac{\rm d}{{\rm d}x}}v \left ( x \right ) =w-B-kv \left ( x \right ) \end{align*}
with initial conditions \(v \left ( 0 \right ) =0\).
Trying separable ODE.
In canonical form, the ODE is written as \begin{align*} v' &= F(x,v)\\ &= -{\frac{kv+B-w}{m}} \end{align*}
The ODE \( \frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} = -{\frac{kv+B-w}{m}}\), is separable. It can be written as \[ \frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} = f( x) g(v) \] Where \(f(x)=1\) and \(g(v)={\frac{-kv-B+w}{m}}\). Therefore
\[ \frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} ={\frac{-kv-B+w}{m}} \] Hence \begin{align*} \left ({\frac{m}{-kv-B+w}}\right )\mathop{\mathrm{d}v}&= \mathop{\mathrm{d}x}\\ \int \left ({\frac{m}{-kv-B+w}}\right )\mathop{\mathrm{d}v}&= \int \mathop{\mathrm{d}x}\\ -{\frac{m\ln \left ( \left | kv+B-w \right | \right ) }{k}}&=x+C_{{1}}\\ \end{align*}
Solving for \(v\) gives \[ v ={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k \left ( x+C_{{1}} \right ) }{m}}}}-B+w \right ) } \]
Initial conditions are now used to solve for \(C_{{1}}\). Substituting \(x=0\) and \(v=0\) in the above solution gives an equation to solve for the constant of integration. \begin{align*} 0&={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{kC_{{1}}}{m}}}}-B+w \right ) } \end{align*}
Hence \[ C_{{1}} = -{\frac{m\ln \left ( -B+w \right ) }{k}} \] Substituting \(C_1\) found above back in the solution gives \[ v \left ( x \right ) ={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k}{m} \left ( x-{\frac{m\ln \left ( -B+w \right ) }{k}} \right ) }}}-B+w \right ) } \]
The solution \(\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k}{m} \left ( x-{\frac{m\ln \left ( -B+w \right ) }{k}} \right ) }}}-B+w \right ) \) can be simplified to \begin{equation} \setlength{\fboxsep }{2.5\fboxsep }\boxed{v \left ( x \right ) ={\frac{1}{k} \left ( -{{\rm e}^{{\frac{m\ln \left ( -B+w \right ) -xk}{m}}}}-B+w \right ) }} \end{equation}
The ODE to solve is \begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) ={x}^{3} \left ( y \left ( x \right ) -x \right ) ^{2}+{\frac{y \left ( x \right ) }{x}} \end{align*}
Trying Riccati ODE.
In canonical form, the ODE is written as \begin{align*} y' &= F(x,y)\\ &={\frac{{x}^{6}-2\,{x}^{5}y+{x}^{4}{y}^{2}+y}{x}} \end{align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' ={x}^{5}-2\,{x}^{4}y+{x}^{3}{y}^{2}+{\frac{y}{x}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x){y}^{2} \]
Shows that \(f_0(x)={x}^{5}\), \(f_1(x)={\frac{-2\,{x}^{5}+1}{x}}\) and \(f_2(x)={x}^{3}\).
Let \begin{align*} y &= \frac{-u'}{f_2 u} \\ &= \frac{-u'}{u{x}^{3}} \tag{1} \end{align*}
Using the above substitution in the given ODE results (after some simplification) in a second order ODE to solve for \(u(x)\) which is \begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag{2} \end{align*}
But \begin{align*} f_2' &=3\,{x}^{2}\\ f_1 f_2 &= \left ( -2\,{x}^{5}+1 \right ){x}^{2}\\ f_2^2 f_0 &={x}^{11} \end{align*}
Substituting the above terms back in (2) gives \begin{align*}{x}^{3}{\frac{{\rm d}^{2}}{{\rm d}{x}^{2}}}u \left ( x \right ) - \left ( 3\,{x}^{2}+ \left ( -2\,{x}^{5}+1 \right ){x}^{2} \right ){\frac{\rm d}{{\rm d}x}}u \left ( x \right ) +{x}^{11}u \left ( x \right ) &=0 \end{align*}
Solving the above ODE gives
\[ u \left ( x \right ) ={{\rm e}^{-1/5\,{x}^{5}}} \left ({x}^{5}C_{{2}}+C_{{1}} \right ) \] The above shows that \[ u'(x) = -{x}^{4}{{\rm e}^{-1/5\,{x}^{5}}} \left ({x}^{5}C_{{2}}+C_{{1}}-5\,C_{{2}} \right ) \] Hence, using the above in (1) gives the solution \[ y \left ( x \right ) ={\frac{x \left ({x}^{5}C_{{2}}+C_{{1}}-5\,C_{{2}} \right ) }{{x}^{5}C_{{2}}+C_{{1}}}} \] Dividing both numerator and denominator by \(C_2\) gives, after renaming the constant \(\frac{C_1}{C_2}=C_0\) the following
\[ \boxed{y \left ( x \right ) ={\frac{x \left ({x}^{5}+C_{{0}}-5 \right ) }{{x}^{5}+C_{{0}}}}} \]
