Question

cos²x dy/dx + y = tan x ,Solve it.

Answer 1

Answer:

The solution is

$\bf\:y.e^{tanx}=tanx\:e^{tanx}-e^{tanx}+c$

Step-by-step explanation:

Cos²x dy/dx + y = tan x ,Solve it.

$cos^2x\:\frac{dy}{dx}+y=tanx$

$\text{Divide both sides by }cos^2x$

$\frac{dy}{dx}+sec^2x\:y=sec^2x\:tanx$

This is a linear differential equation in y

comparing with

$\frac{dy}{dx}+Py=Q$ we get

$P=sec^2x\:and\:Q=sec^2x\:tanx$

Integrating factor

$=e^{\int\:P\:dx}$

$=e^{\int\:sec^2x\:dx}$

$=e^{tanx}$

The solution is

$y.e^{\int\:P\:dx}=\int\:Q\:e^{\int\:P\:dx}\:dx+c$

$y.e^{tanx}=\int\:sec^2x\:tanx\:e^{tanx}\:dx+c$ ......(1)

Take

$t=tanx$

$\frac{dt}{dx}=sec^2x$

$dt=sec^2x\:dx$

(1) becomes

$y.e^{tanx}=\int\:\:t\:e^{t}\:dt+c$

using integration by parts we get

$y.e^{tanx}=t\:e^t-e^t+c$

The solution is

$\boxed{\bf\:y.e^{tanx}=tanx\:e^{tanx}-e^{tanx}+c}$

Answer 2

Answer:

The result is:

$y\cdot e^{tanx}=tanx\cdot e^{tanx}-e^{tanx}+c$

Step-by-step explanation:

Given equation is:

$cos^2x\dfrac{dy}{dx}+y=tanx$

Hence dividing both side of the equation by $cos^2x$ we get:

$\dfrac{dy}{dx}+\dfrac{y}{cos^2x}=\dfrac{tanx}{cos^2x}$

Since we know the formula:

$\dfrac{1}{cosx}=secx$

So:

$\dfrac{dy}{dx}+sec^2x\,(y)=sec^2x\,tanx\,\,\,\,\,eqn(1)$

This is in the form of first order linear differential equation because of $\dfrac{dy}{dx}$.

Hence the general solution of this equation is:

$\dfrac{dy}{dx}+P(x)y=Q(x)\,\,\,\,\,eqn(2)$

Comparing with eqn(2) with eqn(1) we get:

$P(x)=sec^2x\\Q(x)=sec^2x\,tanx$

Now evaluating it for integrating factor:

I.F=$e^{\int P(x)\,dx}$

$=e^{\int sec^2x\,dx}\\=e^{tanx}$

Hence the solution will be:

$y\times I.F=\displaystyle\int Q(x)\times I.F\,dx\\y\times e^{tanx}=\displaystyle\int sec^2x\,tanx\times e^{tanx}\,dx$

Let:

$sec^2x\,dx=dz$

Now integrating:

$\displaystyle\int z\cdot e^z\,dz$

Applying here integration by parts formula and it is describe below:

$\displaystyle\int (uv)dx=u\displaystyle\int v\,dx-\displaystyle\int [\dfrac{d}{dx}(u)\displaystyle\int v\,dx]dx$

Therefore:

$z\displaystyle\int e^z\,dz-\displaystyle\int [\dfrac{d}{dz}(z)\displaystyle\int e^z\,dz]dz\\=ze^z-\displaystyle\int [1\cdot e^z]dz\\=ze^z-e^z+c$

Substituting back the value of z.

$y\cdot e^{tanx}=tanx\,e^{tanx}-e^{tanx}+c$

Here c denotes the integration constant.