Related Formula
For differential equations where terms are functions of a linear expression like (x+y)$(x+y)$, substitute a new variable t = x+y$t = x+y$ to enable variable separation.
Core Logic
Given the differential equation:
(dy)/(dx) = 2x(x+y)³ - x(x+y) - 1$$\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) - 1$$
Let x+y = t 1 + (dy)/(dx) = (dt)/(dx) (dy)/(dx) = (dt)/(dx) - 1$x+y = t \implies 1 + \frac{dy}{dx} = \frac{dt}{dx} \implies \frac{dy}{dx} = \frac{dt}{dx} - 1$.
Substituting these terms back:
(dt)/(dx) - 1 = 2xt³ - xt - 1$$\frac{dt}{dx} - 1 = 2xt^3 - xt - 1$$
(dt)/(dx) = 2xt³ - xt (dt)/(dx) = xt(2t² - 1)$$\frac{dt}{dx} = 2xt^3 - xt \implies \frac{dt}{dx} = xt(2t^2 - 1)$$
Step 1: Separating Variables and Integrating
Separating variables:
(dt)/(t(2t² - 1)) = x dx$$\frac{dt}{t(2t^2 - 1)} = x \, dx$$
Multiply numerator and denominator by t$t$:
(t dt)/(t²(2t² - 1)) = x dx$$\frac{t \, dt}{t^2(2t^2 - 1)} = x \, dx$$
Let t² = z 2t dt = dz t dt = (dz)/(2)$t^2 = z \implies 2t \, dt = dz \implies t \, dt = \frac{dz}{2}$:
∫ (dz)/(2z(2z-1)) = ∫ x dx ∫ (dz)/(z(2z-1)) = ∫ 2x dx$$\int \frac{dz}{2z(2z-1)} = \int x \, dx \implies \int \frac{dz}{z(2z-1)} = \int 2x \, dx$$
Using partial fractions:
∫ ( (2)/(2z-1) - (1)/(z) ) dz = ∫ 2x dx$$\int \left( \frac{2}{2z-1} - \frac{1}{z} \right) dz = \int 2x \, dx$$
ln|2z-1| - ln|z| = x² + C ln|(2z-1)/(z)| = x² + C$$\ln|2z-1| - \ln|z| = x^2 + C \implies \ln\left|\frac{2z-1}{z}\right| = x^2 + C$$
Step 2: Apply the Boundary Condition
Given y(0) = 1$y(0) = 1 \implies$ at x = 0$x = 0$, y = 1 t = 0 + 1 = 1 z = t² = 1$y = 1 \implies t = 0 + 1 = 1 \implies z = t^2 = 1$.
Substituting these value constraints into our integral solution:
ln|(2(1)-1)/(1)| = 0² + C ln(1) = C C = 0$$\ln\left|\frac{2(1)-1}{1}\right| = 0^2 + C \implies \ln(1) = C \implies C = 0$$
Thus:
(2z-1)/(z) = ex² 2 - (1)/(z) = ex² (1)/(z) = 2 - ex² z = 12 - ex²$$\frac{2z-1}{z} = e^{x^2} \implies 2 - \frac{1}{z} = e^{x^2} \implies \frac{1}{z} = 2 - e^{x^2} \implies z = \frac{1}{2 - e^{x^2}}$$
Since z = t² = (x+y)²$z = t^2 = (x+y)^2$, we have:
(x+y)² = 12 - ex²$$(x+y)^2 = \frac{1}{2 - e^{x^2}}$$
Step 3: Evaluate at the Target Value
We need to find the value of the function at x = 1√(2)$x = \frac{1}{\sqrt{2}}$:
( 1√(2) + y( 1√(2)) )² = 12 - e^( 1√(2))² = 12 - e1/2 = 12 - √(e)$$\left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^{2} = \frac{1}{2 - e^{\left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{1}{2 - e^{1/2}} = \frac{1}{2 - \sqrt{e}}$$
Pattern Recognition
Sees: Differential equation format y' = f(x+y)$y' = f(x+y)$.
Shortcut: A linear argument (x+y)$(x+y)$ strongly implies substituting t=x+y$t=x+y$. The final expression requested matched the functional template (x+y)²$(x+y)^2$ exactly, saving steps from extracting standalone square roots.
Chapter Mix
Class 12 Mathematics: Differential Equations
