Question

dy/dx=(3x+2y+4)^2
equation redusebial to veriabal separation eqn

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm \: \dfrac{dy}{dx} = {(3x + 2y + 4)}^{2}$

To solve this differential equation, we use method of substitution.

So substitute,

$\rm \: 3x + 2y + 4 = z$

On differentiating both sides w. r. t. x, we get

$\rm \: 3 + 2\dfrac{dy}{dx} = \dfrac{dz}{dx}$

$\rm \: 2\dfrac{dy}{dx} = \dfrac{dz}{dx} - 3$

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{2}\bigg( \dfrac{dz}{dx} - 3\bigg)$

So, on substituting the values in given differential equation, we get

$\rm \: \dfrac{1}{2}\bigg( \dfrac{dz}{dx} - 3\bigg) = {z}^{2}$

$\rm \: \dfrac{dz}{dx} - 3 = 2{z}^{2}$

$\rm \: \dfrac{dz}{dx} = 2{z}^{2} + 3$

On separating the variables, we get

$\rm \: \dfrac{dz}{ {2z}^{2} + 3} = dx$

On integrating both sides, we get

$\rm \:\displaystyle\int\rm \dfrac{dz}{ {2z}^{2} + 3} = \displaystyle\int\rm dx$

$\rm \: \dfrac{1}{2} \displaystyle\int\rm \dfrac{dz}{ {z}^{2} + \dfrac{3}{2} } = \displaystyle\int\rm dx$

$\rm \: \dfrac{1}{2} \displaystyle\int\rm \dfrac{dz}{ {z}^{2} + {\bigg[\dfrac{ \sqrt{3} }{ \sqrt{2} } \bigg]}^{2} } = x + c$

We know,

$\boxed{\rm{  \:\displaystyle\int\rm \frac{dx}{ {x}^{2} + {a}^{2} } \: = \: \frac{1}{a} {tan}^{ - 1} \frac{x}{a} + c \: \: }}$

So, using this result, we get

$\rm \: \dfrac{1}{2} \times \dfrac{1}{\dfrac{ \sqrt{3} }{ \sqrt{2} } } {tan}^{ - 1} \dfrac{ \sqrt{2}z}{ \sqrt{3} } = x + c$

$\rm \: \dfrac{1}{ \sqrt{6} } {tan}^{ - 1} \dfrac{ \sqrt{2} \: z}{ \sqrt{3} } = x + c$

$\rm \: \dfrac{1}{ \sqrt{6} } {tan}^{ - 1} \dfrac{ \sqrt{2} \: (3x + 2y + 4)}{ \sqrt{3} } = x + c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$