Find a particular solution satisfying the giv | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given differential equation is:$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x$Factorizing the denominator: $(x^{2}(x+1) + 1(x+1)) \frac{dy}{dx} =

Vedclass · Accepted Answer

$y = \frac{1}{4} \log(x+1)^{2}(x^{2}+1)^{3} - \frac{1}{2} 	an^{-1} x + 1$

Vedclass · Answer

(A) The given differential equation is:$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x$Factorizing the denominator: $(x^{2}(x+1) + 1(x+1)) \frac{dy}{dx} = 2x^{2}+x$$(x+1)(x^{2}+1) \frac{dy}{dx} = 2x^{2}+x$$dy = \frac{2x^{2}+x}{(x+1)(x^{2}+1)} dx$Using partial fractions: $\frac{2x^{2}+x}{(x+1)(x^{2}+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+1}$$2x^{2}+x = A(x^{2}+1) + (Bx+C)(x+1)$Setting $x = -1$: $2(-1)^{2} + (-1) = A((-1)^{2}+1) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2}$Comparing coefficients of $x^{2}$: $A+B = 2 \Rightarrow \frac{1}{2} + B = 2 \Rightarrow B = \frac{3}{2}$Comparing constants: $A+C = 0 \Rightarrow \frac{1}{2} + C = 0 \Rightarrow C = -\frac{1}{2}$Integrating: $\int dy = \int \frac{1/2}{x+1} dx + \int \frac{3/2x - 1/2}{x^{2}+1} dx$$y = \frac{1}{2} \log|x+1| + \frac{3}{4} \log(x^{2}+1) - \frac{1}{2} 	an^{-1} x + C$$y = \frac{1}{4} [2 \log|x+1| + 3 \log(x^{2}+1)] - \frac{1}{2} 	an^{-1} x + C$$y = \frac{1}{4} \log[(x+1)^{2}(x^{2}+1)^{3}] - \frac{1}{2} 	an^{-1} x + C$Given $y=1$ when $x=0$: $1 = \frac{1}{4} \log(1) - 0 + C \Rightarrow C = 1$Thus,$y = \frac{1}{4} \log[(x+1)^{2}(x^{2}+1)^{3}] - \frac{1}{2} 	an^{-1} x + 1$

Find a particular solution satisfying the given condition:
$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x; y=1$ when $x=0$

Similar Questions

The differential equation $\cot y \, dx = x \, dy$ has a solution of the form

The solution of the differential equation $2x \frac{dy}{dx} - y = 0$,with the initial condition $y(1) = 2$,represents which of the following curves?

The solution of the differential equation $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ is

If $\frac{a}{a_1}=\frac{b}{b_1}$,then the substitution to be used to solve the differential equation $\frac{dy}{dx}=\frac{ax+by+c}{a_1x+b_1y+c_1}$ by using separation of variables is

If $y=y(x)$ and $\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x$,with $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find a particular solution satisfying the given condition:$(x^{3}+x^{2}+x+1) \frac{dy}{dx} = 2x^{2}+x; y=1$ when $x=0$